The equivalent resistance between the points $A$ and $B$ in the following circuit is $\frac{x}{5} \Omega$. The value of $x$ is $\_\_\_\_$ .
Explanation:
The given circuit is a symmetric unbalanced bridge.
To find the equivalent resistance ( $R_{A B}$ ), we apply a potential difference $V$ across terminals $A$ and $B$ and determine the total current I entering the circuit.
Let the potential at node A be V (input terminal) and the potential at node B be 0.
Using Kirchhoff's Current Law (KCL) the sum of currents leaving each junction is zero.
At node $\mathrm{V}_1$ :
$ \frac{\left(V_1-V\right)}{6}+\frac{\left(V_1-0\right)}{3}+\frac{\left(V_1-V_2\right)}{3}=0 $
$\Rightarrow $ $\left(V_1-V\right)+2 V_1+2\left(V_1-V_2\right)=0$
$\Rightarrow $ $ 5 V_1-2 V_2=V \ldots \text { (i) } $
At node $\mathrm{V}_2$ :
$\frac{V_2-V}{3}+\frac{V_2-0}{6}+\frac{V_2-V_1}{3}=0$
$\Rightarrow $ $2\left(V_2-V\right)+V_2+2\left(V_2-V_1\right)=0$
$\Rightarrow $ $ -2 \mathrm{~V}_1+5 \mathrm{~V}_2=2 \mathrm{~V} \ldots \text { (ii) } $
Solving both the equations, (i) $\times 5+$ (ii) $\times 2$
$ \left(5 V_1-2 V_2\right) \times 5+\left(-2 V_1+5 V_2\right) \times 2=V \times 5+2 V \times 2 $
$\Rightarrow $ $\left(25 V_1-10 V_2\right)+\left(-4 V_1+10 V_2\right)=5 V+4 V$
$\Rightarrow $ $21 V_1=9 V \Rightarrow V_1=\frac{9}{21} V=\frac{3}{7} V$
Substituting $V_1$ into Equation (i) to find $V_2$ :
$ \begin{gathered} 5\left(\frac{3}{7} \mathrm{~V}\right)-2 \mathrm{~V}_2=\mathrm{V} \\ 2 \mathrm{~V}_2=\frac{15}{7} \mathrm{~V}-\mathrm{V}=\frac{8}{7} \mathrm{~V} \Rightarrow \mathrm{~V}_2=\frac{4}{7} \mathrm{~V} \end{gathered} $
The total current I entering from terminal A is the sum of currents in the two left branches :
$ I=\frac{V-V_1}{6}+\frac{V-V_2}{3} $
$\Rightarrow $ $I=\frac{V-\frac{3}{7} V}{6}+\frac{V-\frac{4}{7} V}{3}$
$\Rightarrow $ $I=\frac{\frac{4 V}{7}}{6}+\frac{\frac{3 V}{7}}{3}=\frac{2 V}{21}+\frac{V}{7}$
$ I=\frac{2 V+3 V}{21}=\frac{5 V}{21} $
The equivalent resistance $R_{A B}$ is :
$ R_{A B}=\frac{V}{I}=\frac{V}{5 V / 21}=\frac{21}{5} \Omega $
Given that $\mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{x}}{5} \Omega :$
$ \frac{x}{5}=\frac{21}{5} \Rightarrow x=21 $
Therefore, the value of x is 21.
In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2 \Omega$ and $3 \Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l \mathrm{~cm}$ from the left. Now when an unknown resistance $x \Omega$ is connected in parallel to $3 \Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance $x$ is
$\_\_\_\_$ $\Omega$.
Explanation:
A meter bridge works on the principle of a balanced Wheatstone bridge. The ratio of the resistances in the gaps is equal to the ratio of the lengths of the wire segments.
Case 1 :
Null point from left is l cm . So, the remaining length from right (100 - l) cm
At balanced condition
$ \frac{2}{3}=\frac{l}{100-l} \Rightarrow 2(100-l)=3l $
$\Rightarrow 200-2l=3l$
$\Rightarrow 5 \mathrm{l}=200 \Rightarrow \mathrm{l}=40 \mathrm{~cm}$
Case 2 :
When an unknown resistance x is connected in parallel with the $3 \Omega$ resistor in the right gap. This changes the equivalent resistance of the right gap.
New right gap resistance $S^{\prime}=\frac{3 x}{3+x}$
New null point $\mathrm{l}^{\prime}=\mathrm{l}+10=40+10=50 \mathrm{~cm}$
So, the remaining length from right $=100-50=50 \mathrm{~cm}$
At balanced condition for the second case:
$ \frac{\mathrm{R}}{\mathrm{~S}^{\prime}}=\frac{\mathrm{l}^{\prime}}{100-\mathrm{l}^{\prime}} $
$\Rightarrow $ $\frac{2}{\left(\frac{3 x}{3+x}\right)}=\frac{50}{50} \Rightarrow \frac{6+2 x}{3 x}=1$
$ \Rightarrow 3 x=6+2 x \Rightarrow x=6 \Omega $
The value of unknown resistance x is $6 \Omega$.
Hence, the correct answer is 6.
A cylindrical conductor of length 2 m and area of cross-section $0.2 \mathrm{~mm}^2$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \mathrm{~m}^2 / \mathrm{V} . \mathrm{s}$. The value of $\alpha$ is :
(electron concentration $=5 \times 10^{28} / \mathrm{m}^3$ and electron charge $=1.6 \times 10^{-19} \mathrm{C}$ )
Explanation:
The mobility $(\mu)$ of free electrons in a conductor is defined as the magnitude of their drift velocity $\left(\mathrm{v}_{\mathrm{d}}\right)$ per unit electric field (E).
$ \mu=\frac{v_d}{E} $
The electric field ( E ) established across a conductor of length L connected to a potential difference V is given by:
$ \mathrm{E}=\frac{\mathrm{V}}{\mathrm{~L}} $
The electric current (I) flowing through a conductor with cross-sectional area A and free electron number density n is:
$ \mathrm{I}=\mathrm{neA} \mathrm{v}_{\mathrm{d}} $
$ \Rightarrow \mathrm{v}_{\mathrm{d}}=\frac{1}{\mathrm{ne} \mathrm{~A}} $
Using mobility formula,
$\mu=\frac{\left(\frac{\mathrm{I}}{\mathrm{ne} \mathrm{A}}\right)}{\left(\frac{\mathrm{V}}{\mathrm{L}}\right)}$
$\Rightarrow $ $ \mu=\frac{\mathrm{IL}}{\mathrm{ne} \mathrm{AV}} $
We have given,
Length, $\mathrm{L}=2 \mathrm{~m}$
Current, $\mathrm{I}=1.6 \mathrm{~A}$
Voltage, $\mathrm{V}=2 \mathrm{~V}$
Electron concentration, $\mathrm{n}=5 \times 10^{28} \mathrm{~m}^{-3}$
Electron charge, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$
Area of cross-section, $\mathrm{A}=0.2 \mathrm{~mm}^2=0.2 \times 10^{-6} \mathrm{~m}^2=2 \times 10^{-7} \mathrm{~m}^2$
Substituting the values,
$ \mu=\frac{1.6 \times 2}{\left(5 \times 10^{28}\right) \times\left(1.6 \times 10^{-19}\right) \times\left(2 \times 10^{-7}\right) \times 2} $
$\Rightarrow $ $\mu=\frac{3.2}{32 \times 10^2}$
$\Rightarrow $ $\mu=0.001 \mathrm{~m}^2 / \mathrm{V} \cdot \mathrm{s}$
$\Rightarrow $ $\mu=1 \times 10^{-3} \mathrm{~m}^2 / \mathrm{V} \cdot \mathrm{s}$
$\Rightarrow 1 \times 10^{-3}=\alpha \times 10^{-3} \Rightarrow \alpha=1$
The heat generated in 1 minute between points $A$ and $B$ in the given circuit, when a battery of 9 V with internal resistance of $1 \Omega$ is connected across these points is $\_\_\_\_$ J.
Explanation:
A bridge is balanced if the ratio of the resistances in the two arms is equal :
$ \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{R}_3}{\mathrm{R}_4} $
So, from the given circuit, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{2}$ and $\frac{\mathrm{R}_3}{\mathrm{R}_4}=\frac{2}{4}=\frac{1}{2} \Rightarrow \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{R}_3}{\mathrm{R}_4}$
Since the ratio is equal $\left(\frac{1}{2}=\frac{1}{2}\right)$, the bridge is balanced.
Because the bridge is balanced, no current flows through the central $1 \Omega$ resistor. The circuit now simplifies to two parallel branches :
Here, $\mathrm{R}_1$ and $\mathrm{R}_3$ are in series, $\mathrm{R}_{13}=1+2=3 \Omega$
Also, $\mathrm{R}_2$ and $\mathrm{R}_4$ are in series, $\mathrm{R}_{24}=2+4=6 \Omega$
Now, $\mathrm{R}_{13}$ and $\mathrm{R}_{24}$ are in parallel.
So, the equivalent resistance between A and $\mathrm{B}\left(\mathrm{R}_{\mathrm{AB}}\right)$ is :
$ \frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{1}{2} $
$\Rightarrow $ $ \mathrm{R}_{\mathrm{AB}}=2 \Omega $
The battery has an emf of 9 V and an internal resistance (r) of $1 \Omega$. The total resistance ( $\mathrm{R}_{\text {total }}$ ) is:
$ \mathrm{R}_{\text {total }}=\mathrm{R}_{\mathrm{AB}}+\mathrm{r}=2+1=3 \Omega $
Using Ohm's Law for the whole circuit:
$ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\text {total }}}=\frac{9}{3}=3 \text { Amperes } $
So, the power dissipated between points A and B is,
$\mathrm{P}_{\mathrm{AB}}=\mathrm{I}^2 \cdot \mathrm{R}_{\mathrm{AB}} $
$\Rightarrow $ $ \mathrm{P}_{\mathrm{AB}}=(3)^2 \times 2=9 \times 2=18 \text { Watts } $
So, the total heat $(\mathrm{H})$ generated in $\mathrm{t}=1$ minute $=60$ seconds is :
$ \mathrm{H}=\mathrm{P}_{\mathrm{AB}} \times \mathrm{t} $
$\Rightarrow $ $ \mathrm{H}=18 \times 60=1080 \mathrm{~J} $
Therefore, the heat generated in one minute is 1080 J between points A and B .
Hence, the correct answer is 1080.
Two cells of emfs 1 V and 2 V and internal resistances $2 \Omega$ and $1 \Omega$, respectively, are connected in series with an external resistance of $6 \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is $\mathrm{I}_2$. The value of $\left(\frac{\mathrm{I}_1}{\mathrm{I}_2}\right)$ is $\frac{x}{3}$. The value of $x$ is___________.
Explanation:
Series Connection:
Total emf = 1 V + 2 V = 3 V.
Total resistance = 2 Ω + 1 Ω + 6 Ω = 9 Ω.
Current $I_1 = \frac{\text{emf}}{\text{resistance}} = \frac{3}{9} = \tfrac{1}{3}\text{ A}.$
Parallel Connection:
Equivalent emf: $\varepsilon_{\text{eq}} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\tfrac{1}{2} + \tfrac{2}{1}}{\tfrac{1}{2} + 1} = \frac{5}{3}\text{ V}.$
Equivalent internal resistance of the cells: $r_{\text{int,eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\,\Omega.$
Total resistance = external 6 Ω + internal $\tfrac{2}{3}$ Ω = $\tfrac{20}{3}$ Ω.
Current $I_2 = \frac{\varepsilon_{\text{eq}}}{\text{total resistance}} = \frac{5/3}{20/3} = \tfrac{1}{4}\text{ A}.$
Finally, $\frac{I_1}{I_2} = \frac{\tfrac{1}{3}}{\tfrac{1}{4}} = \frac{4}{3},$ so $x = 4\,$.
In the figure shown below, a resistance of $150.4 \Omega$ is connected in series to an ammeter A of resistance $240 \Omega$. A shunt resistance of $10 \Omega$ is connected in parallel with the ammeter. The reading of the ammeter is___________mA .
Explanation:
Step 1: Find the Total Resistance
First, add up the resistor in series ($R_1 = 150.4~\Omega$) with the combined resistance of the ammeter and its shunt ($R_2$).
The ammeter ($240~\Omega$) is in parallel with a shunt of ($10~\Omega$). The combined resistance of resistors in parallel is found by:
$ R_2 = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6~\Omega $
The total resistance ($R_{eq}$) is:
$ R_{eq} = 150.4 + 9.6 = 160~\Omega $
Step 2: Find the Current in the Circuit
Suppose the total current in the circuit is $I$. But to find the ammeter reading, we care about the current through the ammeter coil itself.
The current through the ammeter, $I_1$, is given by how the current splits between the ammeter and the shunt. The fraction going through the ammeter is:
$ I_1 = \frac{R_2}{240} \times I $
But $R_2 = 9.6~\Omega$ (the parallel combination), and $240~\Omega$ is the ammeter's resistance:
$ I_1 = \frac{9.6}{240} \times I $
Step 3: Substitute the Circuit Values
The supply voltage is $20~V$. The total resistance is $160~\Omega$. So the total current is:
$ I = \frac{20}{160} = 0.125~A $
So: $ I_1 = 0.125 \times \frac{9.6}{240} $
Calculate this: $ I_1 = 0.125 \times 0.04 = 0.005~A = 5~\text{mA} $
Final Answer:
The ammeter reads $5~\text{mA}$.
The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at B, will be _______ A.
Explanation:
$\begin{aligned} & \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \Rightarrow \text { the bridge is balanced } \\ & \Rightarrow \frac{10}{\mathrm{R}}=\frac{20}{40} \\ & \mathrm{R}=20 \Omega \\ & \mathrm{I}=\frac{40}{20}=2 \mathrm{~A} \end{aligned}$
A wire of resistance $9 \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be _________ ohm.
Explanation:
$ \text{Each side of the triangle has a resistance of } R_{\text{side}} = \frac{9\,\Omega}{3} = 3\,\Omega. $
When measuring the equivalent resistance between any two vertices, there are two paths:
A direct path along one side with resistance:
$ R_1 = 3\,\Omega. $
An indirect path passing through the other two sides in series:
$ R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega. $
These two paths are in parallel, so the equivalent resistance is calculated by:
$ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3\,\Omega} + \frac{1}{6\,\Omega} = \frac{2}{6\,\Omega} + \frac{1}{6\,\Omega} = \frac{3}{6\,\Omega}. $
Thus,
$ R_{\text{eq}} = \frac{6\,\Omega}{3} = 2\,\Omega. $
The equivalent resistance across any two vertices of the equilateral triangle is therefore:
$ 2\,\Omega. $
The net current flowing in the given circuit is __________ A.
Explanation:
$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=2 \Omega \\ & \mathrm{I}=\frac{2}{2}=1 \mathrm{~A} \end{aligned}$
To determine the resistance (R) of a wire, a circuit is designed below. The $V$-$I$ characteristic curve for this circuit is plotted for the voltmeter and the ammeter readings as shown in figure. The value of $R$ is _________ $\Omega$.
Explanation:
$\begin{aligned} & R_{\text {net }}=\frac{1}{\text { Slope }}=\frac{8}{4} \times 10^3=2 \times 10^3 \\ & \Rightarrow 2=\frac{10 R}{10+R} \\ & \Rightarrow 20+2 R=10 R \\ & \Rightarrow 8 R=20 \\ & \Rightarrow R=2.5 \mathrm{k} \Omega \\ & \quad=2500 ~\Omega \end{aligned}$
At room temperature $(27^{\circ} \mathrm{C})$, the resistance of a heating element is $50 \Omega$. The temperature coefficient of the material is $2.4 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$. The temperature of the element, when its resistance is $62 \Omega$, is __________${ }^{\circ} \mathrm{C}$.
Explanation:
We can start solving this problem by first understanding that the resistance of a material changes with temperature, and this change can be quantified using the temperature coefficient of resistance $ \alpha $. The relationship between the resistance of a material at any temperature $ T $ and its resistance at a reference temperature $ T_0 $ is given by the formula:
$ R = R_0(1 + \alpha(T - T_0)) $
Where:
- $ R $ is the resistance at temperature $ T $ (in this case, $ 62 \Omega $).
- $ R_0 $ is the resistance at reference temperature $ T_0 $ (in this case, $ 50 \Omega $).
- $ \alpha $ is the temperature coefficient of resistance (in this case, $ 2.4 \times 10^{-4} \, ^\circ\mathrm{C}^{-1} $).
- $ T $ is the unknown temperature we need to find.
- $ T_0 $ is the reference temperature, given as $ 27^\circ \mathrm{C} $.
By substituting the given values into the formula, we get:
$ 62 = 50(1 + 2.4 \times 10^{-4}(T - 27)) $
First, divide both sides of the equation by 50:
$ \frac{62}{50} = 1 + 2.4 \times 10^{-4}(T - 27) $
Then solve for $ T $:
$ 1.24 = 1 + 2.4 \times 10^{-4}(T - 27) $
$ 0.24 = 2.4 \times 10^{-4}(T - 27) $
$ \frac{0.24}{2.4 \times 10^{-4}} = T - 27 $
$ 1000 = T - 27 $
$ T = 1027 ^\circ\mathrm{C} $
Thus, the temperature of the element when its resistance is $ 62 \Omega $ is $ 1027^\circ\mathrm{C} $.
The current flowing through the $1 \Omega$ resistor is $\frac{n}{10}$ A. The value of $n$ is _______.
Explanation:
At $C$
$\begin{aligned} & \frac{v_1}{2}+\frac{v_1-5}{2}+\frac{v_1+10-v_2}{1}=0 \\ & v_2=2 v_1+\frac{15}{2} \quad \text{.... (i)} \end{aligned}$
At $A$
$\begin{aligned} & \frac{v_2-5}{4}+\frac{v_2}{4}+\frac{v_2-10-v_1}{1}=0 \\ & 6 v_2=4 v_1+45 \quad \text{.... (ii)} \end{aligned}$
$\begin{aligned} & \Rightarrow v_1=0 \\ & \text { and } v_2=\frac{15}{2} \\ & \therefore \quad i=\frac{5}{2} \mathrm{~A} \\ & \Rightarrow n=25 \end{aligned}$
A heater is designed to operate with a power of $1000 \mathrm{~W}$ in a $100 \mathrm{~V}$ line. It is connected in combination with a resistance of $10 \Omega$ and a resistance $R$, to a $100 \mathrm{~V}$ mains as shown in figure. For the heater to operate at $62.5 \mathrm{~W}$, the value of $\mathrm{R}$ should be _______ $\Omega$.
Explanation:
$\begin{gathered} R_H=\frac{100 \times 100}{1000}=10 \Omega \\ i_H=\sqrt{\frac{62.5}{10}}=2.5 \mathrm{~A} \\ V_H=25 \mathrm{~V} \\ \Rightarrow \quad i_b=\frac{75}{10}=7.5 \mathrm{~A} \\ \Rightarrow \quad i_R=5 \mathrm{~A} \\ R=\frac{25}{5}=5 \Omega \end{gathered}$
Resistance of a wire at $0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C}$ and $t^{\circ} \mathrm{C}$ is found to be $10 \Omega, 10.2 \Omega$ and $10.95 \Omega$ respectively. The temperature $t$ in Kelvin scale is _________.
Explanation:
To determine the temperature $t$ in the Kelvin scale, we need to use the relationship between the resistance of a wire and temperature. The general formula for the resistance $R$ of a wire as a function of temperature is:
$ R_t = R_0 (1 + \alpha t) $
where:
- $R_t$ is the resistance at temperature $t$
- $R_0$ is the resistance at the reference temperature (usually $0^{\circ} \mathrm{C}$)
- $\alpha$ is the temperature coefficient of resistance
- $t$ is the temperature
We are given the following resistances:
- Resistance at $0^{\circ} \mathrm{C}$: $R_0 = 10 \Omega$
- Resistance at $100^{\circ} \mathrm{C}$: $R_{100} = 10.2 \Omega$
- Resistance at $t^{\circ} \mathrm{C}$: $R_t = 10.95 \Omega$
First, we need to find the temperature coefficient of resistance $\alpha$. Using the resistance at $100^{\circ} \mathrm{C}$:
$ 10.2 = 10 (1 + \alpha \cdot 100) $
Solving for $\alpha$:
$ \frac{10.2}{10} = 1 + 100\alpha \implies 1.02 = 1 + 100\alpha \implies 100\alpha = 0.02 \implies \alpha = \frac{0.02}{100} = 0.0002 $
Now, we can find the temperature $t$ using the resistance at $t^{\circ} \mathrm{C}$:
$ 10.95 = 10 (1 + 0.0002 \cdot t) $
Solving for $t$:
$ \frac{10.95}{10} = 1 + 0.0002 t \implies 1.095 = 1 + 0.0002 t \implies 0.0002 t = 0.095 \implies t = \frac{0.095}{0.0002} = 475 $
The temperature $t$ in Celsius is $475^{\circ} \mathrm{C}$. To convert this to the Kelvin scale:
$ T_{K} = t_{C} + 273.15 = 475 + 273.15 = 748.15 \, \mathrm{K} $
So, the temperature $t$ in Kelvin scale is approximately $748.15 \, \mathrm{K}$.
In the given figure an ammeter A consists of a $240 \Omega$ coil connected in parallel to a $10 \Omega$ shunt. The reading of the ammeter is ________ $\mathrm{mA}$.
Explanation:
$\begin{aligned} & i=\frac{24}{140 \cdot 4+r_A}=\frac{24}{140 \cdot 4+9 \cdot 6}=0.16 \mathrm{~A} \\ & i=160 \mathrm{~mA} \end{aligned}$
A wire of resistance $R$ and radius $r$ is stretched till its radius became $r / 2$. If new resistance of the stretched wire is $x ~R$, then value of $x$ is ________.
Explanation:
The resistance $R$ of a wire is given by the formula:
$ R = \rho \frac{l}{A}, $where:
- $\rho$ is the resistivity of the material,
- $l$ is the length of the wire,
- $A$ is the cross-sectional area of the wire.
If we have a cylindrical wire, the cross-sectional area can be expressed as $A = \pi r^2$, where $r$ is the radius of the cylinder. Therefore, the resistance of the original wire can be written as:
$ R = \rho \frac{l}{\pi r^2}. $When the wire is stretched such that its radius becomes $r / 2$, its volume would remain constant, given that the volume of a cylinder is $V = A \cdot l = \pi r^2 \cdot l$. Assuming the volume before and after the stretching is the same, and since the area is now a quarter of the original (because when the radius is halved, the area, which is proportional to the square of the radius, is reduced to a quarter), the length must have increased to four times the original to preserve the volume. That is,
$ l' = 4l, $and the new area,
$ A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}. $Therefore, the new resistance, $R'$, of the wire can be calculated using the original formula for resistance:
$ R' = \rho \frac{l'}{A'} = \rho \frac{4l}{\frac{\pi r^2}{4}} = \rho \frac{4l}{\pi r^2} \cdot 4 = 16 \rho \frac{l}{\pi r^2} = 16R. $Hence, the new resistance of the wire is $16R$, which means $x = 16$.
A wire of resistance $20 \Omega$ is divided into 10 equal parts, resulting pairs. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _________ $\Omega$.
Explanation:
Let's start by understanding the process of dividing the wire and recombining its parts to form the final configuration. Initially, we have a wire with a resistance of $20 \Omega$. This wire is divided into 10 equal parts, each part then has a resistance of:
$\frac{20 \Omega}{10} = 2 \Omega$
Since each part has the same length and presumably the same material and cross-sectional area, then each part will have the same resistance of $2 \Omega$.
When two parts are connected in parallel, the equivalent resistance, $R_{\text{parallel}}$, of this configuration can be calculated using the formula for two resistors in parallel:
$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}$
Given that $R_1 = R_2 = 2 \Omega$ (since the parts are identical), we have:
$\frac{1}{R_{\text{parallel}}} = \frac{1}{2 \Omega} + \frac{1}{2 \Omega} = \frac{2}{2 \Omega}$
This simplifies to:
$\frac{1}{R_{\text{parallel}}} = \frac{2}{2 \Omega} = \frac{1}{\Omega}$
From which it follows that:
$R_{\text{parallel}} = 1 \Omega$
Now, since the original wire was divided into 10 equal parts, and pairs of these parts are connected in parallel, this results in $\frac{10}{2} = 5$ pairs. Each of these pairs has an equivalent resistance of $1 \Omega$.
Finally, these pairs are all connected in series. The total resistance of resistors in series is simply the sum of their individual resistances. Therefore, the equivalent resistance of the final configuration, $R_{\text{series}}$, is:
$R_{\text{series}} = 5 \times R_{\text{parallel}} = 5 \times 1 \Omega = 5 \Omega$
So, the equivalent resistance of the final combination is $5 \Omega$.
In the experiment to determine the galvanometer resistance by half-deflection method, the plot of $1 / \theta$ vs the resistance (R) of the resistance box is shown in the figure. The figure of merit of the galvanometer is _________ $\times 10^{-1} \mathrm{~A} /$ division. [The source has emf $2 \mathrm{~V}$]
Explanation:
$\frac{1}{3} \mathrm{~A} \longrightarrow \frac{1}{2} \mathrm{div}$
$\frac{1}{2} \mathrm{~A} \longrightarrow \frac{2}{3} \mathrm{div}$
$\text { Figure of merit }=\frac{\Delta i}{\Delta \theta} \quad \begin{aligned} & \frac{\frac{1}{2}--}{\frac{2}{3}-\frac{1}{2}} \\ &=0.5 \\ &=5 \times 10^{-1} \mathrm{~A} / \mathrm{div} \end{aligned}$
Two wires $A$ and $B$ are made up of the same material and have the same mass. Wire $A$ has radius of $2.0 \mathrm{~mm}$ and wire $B$ has radius of $4.0 \mathrm{~mm}$. The resistance of wire $B$ is $2 \Omega$. The resistance of wire $A$ is ________ $\Omega$.
Explanation:
$\begin{aligned} & R=\rho \frac{I}{A}=\rho \frac{V}{A^2} \\ & \text { and } \pi r_1^2 I_1=\pi r_2^2 I_2 \\ & A_1 I_1=A_2 I_2 \\ & \text { So } \frac{R_1}{R_2}=\left(\frac{A_2}{A_1}\right)^2 \\ & \Rightarrow \frac{R}{2}=\left(\frac{r_2}{r_1}\right)^4 \\ & \Rightarrow R=32 \end{aligned}$
Twelve wires each having resistance $2 \Omega$ are joined to form a cube. A battery of $6 \mathrm{~V}$ emf is joined across point $a$ and $c$. The voltage difference between $e$ and $f$ is ________ V.
Explanation:
The circuit can be simplified as
$\begin{aligned} & R_{a c}=\frac{6 \times 2}{8}=\frac{3}{2} \Omega \\ & \begin{aligned} & i=1 \mathrm{Amp} . \\ & V_{e f}=\left(\frac{i}{2}\right)^2 \\ & \quad=1 \mathrm{~V} \end{aligned} \end{aligned}$
Explanation:
To find the amount of electric charge that flows through a section of the conductor, we have to integrate the current over the given time interval. The current $I(t)$ as a function of time $t$ is given by:
$ I=3t^2+4t^3 $The electric charge $Q$ that flows through the conductor from time $t = 1$ s to $t = 2$ s is calculated by integrating the current $I(t)$ with respect to time over this interval:
$ Q = \int_{t_1}^{t_2} I(t) \, dt $Substituting the given limits ($t_1=1$ and $t_2=2$) and the expression for $I(t)$, we get:
$ Q = \int_{1}^{2} (3t^2+4t^3) \, dt $Now we'll integrate the function with respect to $t$:
$ Q = \left[ \frac{3}{3}t^3 + \frac{4}{4}t^4 \right]_{1}^{2} $Simplifying the integrated function:
$ Q = \left[ t^3 + t^4 \right]_{1}^{2} $Substitute the upper and lower limits of the integration:
$ Q = \left[ (2)^3 + (2)^4 \right] - \left[ (1)^3 + (1)^4 \right] $ $ Q = \left[ 8 + 16 \right] - \left[ 1 + 1 \right] $ $ Q = 24 - 2 $ $ Q = 22 \text{ C} $Therefore, the amount of electric charge that flows through the section of the conductor from $t=1$ s to $t=2$ s is 22 Coulombs.
In the following circuit, the battery has an emf of $2 \mathrm{~V}$ and an internal resistance of $\frac{2}{3} \Omega$. The power consumption in the entire circuit is _________ W.
Explanation:
$\begin{aligned} & \mathrm{R}_{\text {eq }}=\frac{4}{3} \Omega \\ & \therefore \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}_{\text {eq }}}=\frac{4}{4 / 3}=3 \mathrm{~W} \end{aligned}$
Equivalent resistance of the following network is __________ $\Omega$.
Explanation:
$6\Omega$ is short circuit
$\mathrm{R}_{\mathrm{eq}}=3 \times \frac{1}{3}=1 \Omega$
Two resistance of $100 \Omega$ and $200 \Omega$ are connected in series with a battery of $4 \mathrm{~V}$ and negligible internal resistance. A voltmeter is used to measure voltage across $100 \Omega$ resistance, which gives reading as $1 \mathrm{~V}$. The resistance of voltmeter must be _______ $\Omega$.
Explanation:
$\begin{aligned} & \frac{R_v 100}{R_v+100}=\frac{200}{3} \\ & 3 R_v=2 R_v+200 \\ & R_v=200 \end{aligned}$
Two cells are connected in opposition as shown. Cell $\mathrm{E}_1$ is of $8 \mathrm{~V}$ emf and $2 \Omega$ internal resistance; the cell $\mathrm{E}_2$ is of $2 \mathrm{~V}$ emf and $4 \Omega$ internal resistance. The terminal potential difference of cell $\mathrm{E}_2$ is __________ V.
Explanation:
$I=\frac{8-2}{2+4}=\frac{6}{6}=1 \mathrm{~A}$
Applying Kirchhoff from C to B
$\begin{aligned} & V_C-2-4 \times 1=V_B \\\\ & V_C-V_B=6 \mathrm{~V} \\\\ & =6 \mathrm{~V} \end{aligned}$
In the given circuit, the current flowing through the resistance $20 \Omega$ is $0.3 \mathrm{~A}$, while the ammeter reads $0.9 \mathrm{~A}$. The value of $\mathrm{R}_1$ is _________ $\Omega$.
Explanation:
Given, $\mathrm{i}_1=0.3 \mathrm{~A}, \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=0.9 \mathrm{~A}$
So, $\mathrm{V}_{\mathrm{AB}}=\mathrm{i}_1 \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}$
$\begin{aligned} & \mathrm{i}_2=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A} \\ & \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=\frac{9}{10} \mathrm{~A} \\ & \frac{3}{10}+\frac{2}{5}+\mathrm{i}_3=\frac{9}{10} \\ & \frac{7}{10}+\mathrm{i}_3=\frac{9}{10} \\ & \mathrm{i}_3=0.2 \mathrm{~A} \\ & \mathrm{So}, \mathrm{i}_3 \times \mathrm{R}_1=6 \mathrm{~V} \\ & (0.2) \mathrm{R}_1=6 \\ & \mathrm{R}_1=\frac{6}{0.2}=30 \Omega \end{aligned}$
Explanation:
In the circuit $I=\frac{9}{3}=3 \mathrm{~A}$
$ \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}=2 \times 1.5=3~~......(I) $
$ \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=4 \times 1.5=6~~......(II) $
$\mathrm{Eq}^{\mathrm{n}}(\mathrm{II})-\mathrm{Eq}^{\mathrm{n}}(\mathrm{I})$
$ \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-3=3 \text { Volt } $
When a resistance of $5 ~\Omega$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $250 \mathrm{~mA}$, however when $1050 ~\Omega$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ $\Omega$.
Explanation:
$\frac{250 \ \text{mA} \times 5}{5 + R_G} = i$
$i = \frac{25}{1050 + R_G}$
Equating the two expressions for current, $i$:
$\frac{250 \ \text{mA} \times 5}{5 + R_G} = \frac{25}{1050 + R_G}$
This equation simplifies to:
$100(5 + R_G) = 1050 \times 5 + R_G \times 5$
Solving for the resistance of the galvanometer, $R_G$:
$95 R_G = 4750$
$R_G = 50 \ \Omega$
So, the resistance of the galvanometer is $50 \ \Omega$.
A potential $\mathrm{V}_{0}$ is applied across a uniform wire of resistance $R$. The power dissipation is $P_{1}$. The wire is then cut into two equal halves and a potential of $V_{0}$ is applied across the length of each half. The total power dissipation across two wires is $P_{2}$. The ratio $P_{2}: \mathrm{P}_{1}$ is $\sqrt{x}: 1$. The value of $x$ is ___________.
Explanation:
The power dissipation, $P_1$, can be calculated using the formula:
$P_1 = \frac{V_0^2}{R}$
Now, let's consider the case where the wire is cut into two equal halves. Each half will have half the original resistance, $\frac{R}{2}$. The potential $V_0$ is applied across the length of each half.
For each half of the wire, the power dissipation, $P'$, can be calculated using the formula:
$P' = \frac{V_0^2}{\frac{R}{2}} = \frac{2V_0^2}{R}$
Since there are two halves of the wire, the total power dissipation across the two wires, $P_2$, is:
$P_2 = 2P' = 2\left(\frac{2V_0^2}{R}\right) = \frac{4V_0^2}{R}$
Now, let's find the ratio $P_2 : P_1$:
$\frac{P_2}{P_1} = \frac{\frac{4V_0^2}{R}}{\frac{V_0^2}{R}} = 4$
Comparing this to the given ratio $\sqrt{x} : 1$, we have:
$\frac{P_2}{P_1} = \sqrt{x}$
So, $\sqrt{x} = 4$.
Squaring both sides, we get:
$x = 16$
The value of $x$ is 16.
The current flowing through a conductor connected across a source is $2 \mathrm{~A}$ and 1.2 $\mathrm{A}$ at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ respectively. The current flowing through the conductor at $50^{\circ} \mathrm{C}$ will be ___________ $\times 10^{2} \mathrm{~mA}$.
Explanation:
First, you establish a relationship between the currents and resistances at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$:
$i_0 R_0 = i_{100} R_{100}$
Plugging in the given values for $i_0$ and $i_{100}$:
$2 R_0 = 1.2 R_0 (1 + 100\alpha) ~\cdots (1)$
From this equation, you find the value of $\alpha$:
$1 + 100\alpha = \frac{5}{3} \Rightarrow 100\alpha = \frac{2}{3} \Rightarrow 50\alpha = \frac{1}{3}$
Now, you need to find the current $i_{50}$ at $50^{\circ} \mathrm{C}$. To do this, you calculate the resistance $R_{50}$ using the found value of $\alpha$:
$R_{50} = R_0 (1 + 50\alpha) = R_0 (1 + \frac{1}{3})$
Using the fact that the voltage across the conductor remains constant, you can find the current $i_{50}$:
$i_{50} = \frac{i_0 R_0}{R_{50}} = \frac{2 \times R_0}{R_0 (1 + \frac{1}{3})} = \frac{2}{1 + \frac{1}{3}} = 1.5 \mathrm{~A}$
Thus, the current flowing through the conductor at $50^{\circ} \mathrm{C}$ is $15 \times 10^2 \mathrm{~mA}$
Two identical cells each of emf $1.5 \mathrm{~V}$ are connected in series across a $10 ~\Omega$ resistance. An ideal voltmeter connected across $10 ~\Omega$ resistance reads $1.5 \mathrm{~V}$. The internal resistance of each cell is __________ $\Omega$.
Explanation:
Since the two cells are connected in series, their internal resistances add up, and the total internal resistance of the series combination is $2r$. The total EMF of the series combination of cells is $1.5\,\text{V} + 1.5\,\text{V} = 3\,\text{V}$.
Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:
$\text{EMF}_{total} - I(R + 2r) = 0$
We are given that the voltage across the $10\,\Omega$ resistor, as measured by the ideal voltmeter, is $1.5\,\text{V}$. According to Ohm's law, the current in the circuit can be determined as:
$I = \frac{V}{R} = \frac{1.5\,\text{V}}{10\,\Omega} = 0.15\,\text{A}$
Now, substitute the given values into the KVL equation:
$3\,\text{V} - 0.15\,\text{A}(10\,\Omega + 2r) = 0$
Solve for $2r$:
$3\,\text{V} - 1.5\,\text{V} = 0.15\,\text{A} \cdot 2r$
$1.5\,\text{V} = 0.3\,\text{A} \cdot r$
Now, solve for the internal resistance $r$:
$r = \frac{1.5\,\text{V}}{0.3\,\text{A}} = 5\,\Omega$
So, the internal resistance of each cell is $5\,\Omega$.
In the circuit diagram shown in figure given below, the current flowing through resistance $3 ~\Omega$ is $\frac{x}{3} A$.
The value of $x$ is ___________
Explanation:
$ \mathrm{E}_2-\mathrm{E}_1=8-4=4 \mathrm{~V} $
$ \begin{aligned} & \frac{1}{3}+\frac{1}{6}=\frac{1}{2}=\frac{1}{R} \\\\ & R=2 \Omega \end{aligned} $
$ I=\frac{4}{8}=0.5 \mathrm{~A} $
$ \begin{aligned} & \mathrm{I}_1=\left(\frac{6}{3+6}\right) \times 0.5 \\\\ & \mathrm{I}_1=\frac{2}{3} \times 0.5=\frac{1}{3} \mathrm{~A} \end{aligned} $
$ I_1=\frac{x}{3}=\frac{1}{3} \therefore x=1 $
A rectangular parallelopiped is measured as $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$. If its specific resistance is $3 \times 10^{-7} ~\Omega \mathrm{m}$, then the resistance between its two opposite rectangular faces will be ___________ $\times 10^{-7} ~\Omega$.
Explanation:
The resistance of a material can be calculated using the formula:
$ R = \rho \frac{L}{A} $
where
- $R$ is the resistance,
- $\rho$ (rho) is the resistivity or specific resistance of the material,
- $L$ is the length (or distance over which the resistance is being measured), and
- $A$ is the cross-sectional area through which the current flows.
In the context of a rectangular parallelepiped, the "length" and "cross-sectional area" can vary depending on which faces of the shape are considered.
In this particular calculation, the resistance is being measured between the two smaller faces of the parallelepiped, which are squares of side length 1 cm:
The cross-sectional area $A$ should be:$A = 100 \, \text{cm} \times 1 \, \text{cm} = 100 \, \text{cm}^2$
Converting this to meters gives:
$A = 100 \, \text{cm}^2 = 1 \, \text{m} \times 0.01 \, \text{m} = 0.01 \, \text{m}^2$
So, if we use these values for the length $L$ and cross-sectional area $A$ in the resistance formula, we get:
$ R = \rho \frac{L}{A} = 3 \times 10^{-7} \Omega \, m \times \frac{0.01 \, m}{0.01 \, \text{m}^2} = 3 \times 10^{-7} \Omega $
Therefore, the resistance between the two smaller faces of the rectangular parallelepiped is $3 \times 10^{-7} \Omega$.
10 resistors each of resistance 10 $\Omega$ can be connected in such as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be ___________.
Explanation:
When resistors are connected in series, the equivalent resistance is the sum of individual resistances. Therefore, the maximum equivalent resistance will be when all 10 resistors are connected in series, giving a total resistance of 100 $\Omega$.
When resistors are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. Therefore, the minimum equivalent resistance will be when all 10 resistors are connected in parallel, giving a total resistance of 1 $\Omega$.
The ratio of maximum to minimum equivalent resistance is thus $\frac{100}{1}=100$.
The number density of free electrons in copper is nearly $8 \times 10^{28} \mathrm{~m}^{-3}$. A copper wire has its area of cross section $=2 \times 10^{-6} \mathrm{~m}^{2}$ and is carrying a current of $3.2 \mathrm{~A}$. The drift speed of the electrons is ___________ $\times 10^{-6} \mathrm{ms}^{-1}$
Explanation:
Using the formula:
$ I = n e A v $
where $I$ is the current, $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $v$ is the drift speed of the electrons.
We can isolate $v$ to find:
$ v = \frac{I}{n e A}$
Substituting the given values:
$ v = \frac{3.2 \, \text{A}}{8 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C} \times 2 \times 10^{-6} \, \text{m}^2}$
This simplifies to:
$ v = \frac{3.2}{16 \times 1.6 \times 10^3} \, \text{ms}^{-1} = 125 \times 10^{-6} \, \text{ms}^{-1} $
So, the drift speed of the electrons is indeed $125 \times 10^{-6} \, \text{ms}^{-1}$.
A current of $2 \mathrm{~A}$ flows through a wire of cross-sectional area $25.0 \mathrm{~mm}^{2}$. The number of free electrons in a cubic meter are $2.0 \times 10^{28}$. The drift velocity of the electrons is __________ $\times 10^{-6} \mathrm{~ms}^{-1}$ (given, charge on electron $=1.6 \times 10^{-19} \mathrm{C}$ ).
Explanation:
The drift velocity $v_d$ can be found using the formula for current $I$ in a conductor:
$I = nqAv_d$,
where:
- $n$ is the number density of free electrons (number of free electrons per unit volume),
- $q$ is the charge of an electron,
- $A$ is the cross-sectional area of the conductor, and
- $v_d$ is the drift velocity of the electrons.
We can rearrange the above formula to solve for $v_d$:
$v_d = \frac{I}{nqA}$.
Given that $I = 2 \, \text{A}$, $n = 2.0 \times 10^{28} \, \text{m}^{-3}$, $q = 1.6 \times 10^{-19} \, \text{C}$, and $A = 25.0 \, \text{mm}^{2} = 25.0 \times 10^{-6} \, \text{m}^{2}$, we can substitute these values into the formula to find $v_d$:
$v_d = \frac{2}{(2.0 \times 10^{28})(1.6 \times 10^{-19})(25.0 \times 10^{-6})}$
$v_d = 25 \times 10^{-6} \, \text{ms}^{-1}$.
Therefore, the drift velocity of the electrons is $25 \times 10^{-6} \, \text{ms}^{-1}$.
As shown in the figure, the voltmeter reads $2 \mathrm{~V}$ across $5 ~\Omega$ resistor. The resistance of the voltmeter is _________ $\Omega$.
Explanation:
$ i=\frac{1 V}{2 \Omega}=\frac{1}{2} A $
$\therefore$ Current through voltmeter $=i-i_1$
$ =\frac{1}{2}-\frac{2}{5}=\frac{5-4}{10}=\frac{1}{10} \mathrm{~A} $
$\therefore$ For voltmeter
$ 2=\left(\frac{1}{10}\right) R \Rightarrow R=20 \Omega $
The length of a metallic wire is increased by $20 \%$ and its area of cross section is reduced by $4 \%$. The percentage change in resistance of the metallic wire is __________.
Explanation:
The resistance ($R$) of a wire can be calculated by the formula:
$ R = \rho \frac{L}{A}, $
where
- $\rho$ is the resistivity (a property of the material),
- $L$ is the length of the wire, and
- $A$ is the cross-sectional area of the wire.
If the length ($L$) is increased by 20%, $L$ becomes $1.2L$.
If the cross-sectional area ($A$) is reduced by 4%, $A$ becomes $0.96A$.
The new resistance $R'$ is then:
$ R' = \rho \frac{1.2L}{0.96A} = 1.25R, $
so the resistance has increased by 25%.
Therefore, the percentage change in the resistance of the metallic wire is 25%.
In the given circuit, the value of $\left| {{{{\mathrm{I_1}} + {\mathrm{I_3}}} \over {{\mathrm{I_2}}}}} \right|$ is _____________
Explanation:
$\begin{aligned} & \mathrm{I}_1=\mathrm{I}_2=\frac{20-10}{10}=1 \mathrm{~A} \\\\ & \mathrm{I}_3=1 \mathrm{~A} \\\\ & \left|\frac{\mathrm{I}_1+\mathrm{I}_3}{\mathrm{I}_2}\right|=2\end{aligned}$
In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5 \mathrm{~V}$ is found to be $60 \mathrm{~cm}$. If this cell is replaced by another cell of emf E, the length-of null point increases by $40 \mathrm{~cm}$. The value of $E$ is $\frac{x}{10} V$. The value of $x$ is ____________.
Explanation:
$E \propto l$
$ \begin{aligned} & \frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}} \\\\ & \frac{1.5}{E}=\frac{60}{100} \\\\ & E=\frac{150}{60}=\frac{5}{2}=\frac{25}{10} \\\\ & \text { so } x=25 \end{aligned} $
Two identical cells, when connected either in parallel or in series gives same current in an external resistance $5 ~\Omega$. The internal resistance of each cell will be ___________ $\Omega$.
Explanation:
$\varepsilon_{\text {series }}=\varepsilon_{1}+\varepsilon_{2}=2 \varepsilon$
$r_{\text {series }}=r_{1}+r_{2}=2 r$
$i=\frac{2 \varepsilon}{5+2 r}$ ......(1)
$\varepsilon_{\text {parallel }}=\frac{\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{r_{2}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}}=\varepsilon$
$ r_{\text {parallel }}=\frac{r}{2}$
$ i=\frac{\varepsilon}{\frac{r}{2}+5} $ .......(2)
Equating (1) and (2), we get
$\Rightarrow \frac{2 \varepsilon}{2 r+5}=\frac{\varepsilon}{\frac{r}{2}+5}$
$\Rightarrow r+10=2 r+5 $
$\Rightarrow r=5 \Omega$
If the potential difference between $\mathrm{B}$ and $\mathrm{D}$ is zero, the value of $x$ is $\frac{1}{n} \Omega$. The value of $n$ is __________.
Explanation:
The circuit is a Wheatstone bridge, so
${{{{6 \times 3} \over {6 + 3}}} \over {{{x \times 1} \over {x + 1}}}} = {{1 + 2} \over x}$
$ \Rightarrow {{2(x + 1)} \over x} = {3 \over x}$
$ \Rightarrow x = {1 \over 2}$
So $n = 2$
In the following circuit, the magnitude of current I1, is ___________ A.
Explanation:
The indicated diagram shows current flow diagram loops for writing Kirchhoff's law are also indicated, writing the equation
$2{I_3} + {I_1} + {I_3} + {I_2} = 5$
or ${I_1} + {I_2} + 3{I_3} = 5$ ..... (1)
${I_2} - 5 = 2({I_3} - {I_2}) + ({I_1} + {I_3} - {I_2})$
or ${I_1} - 4{I_2} + 3{I_3} = - 5$ ...... (2)
$({I_1} + {I_3}) + ({I_1} + {I_3} - {I_2}) = 2$
or $2{I_1} - {I_2} + 2{I_3} = 2$ ...... (3)
on solving ${I_1} = {3 \over 2}A,{I_2} = 2,{I_3} = {1 \over 2}A$
$ = 1.50$
A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5$\Omega$. When a resistance of 15$\Omega$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________$\Omega$.
Explanation:
Let the emf is E and internal resistance is r of this secondary cell so
${{RE} \over {r + R}} \propto l$
so ${{{R_1}E} \over {r + {R_1}}} \propto {l_1}$
& ${{{R_2}E} \over {r + {R_2}}} \propto {l_2}$
$ \Rightarrow {{{R_1}(r + {R_2})} \over {{R_2}(r + {R_1})}} = {{{l_1}} \over {{l_2}}}$
or ${{5(r + 15)} \over {15(r + 5)}} = {{200} \over {300}}$
$ \Rightarrow r = 5\,\Omega $
When two resistance $\mathrm{R_1}$ and $\mathrm{R_2}$ connected in series and introduced into the left gap of a meter bridge and a resistance of 10 $\Omega$ is introduced into the right gap, a null point is found at 60 cm from left side. When $\mathrm{R_1}$ and $\mathrm{R_2}$ are connected in parallel and introduced into the left gap, a resistance of 3 $\Omega$ is introduced into the right gap to get null point at 40 cm from left end. The product of $\mathrm{R_1}$ $\mathrm{R_2}$ is ____________$\Omega^2$
Explanation:
As per given information
${{{R_1} + {R_2}} \over {10}} = {{0.6} \over {0.4}}$ ...... (1)
& ${{{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \over 3} = {{0.4} \over {0.6}}$ ..... (2)
$ \Rightarrow \left. \matrix{ {R_1} + {R_2} = 15 \hfill \cr \& \,{R_1}{R_2} = 30 \hfill \cr} \right] \Rightarrow {R_1}{R_2} = 30\,{\Omega ^2}$
In a metre bridge experiment the balance point is obtained if the gaps are closed by 2$\Omega$ and 3$\Omega$. A shunt of X $\Omega$ is added to 3$\Omega$ resistor to shift the balancing point by 22.5 cm. The value of X is ___________.
Explanation:
$\Rightarrow I=40 \mathrm{~cm}$
as $3 \Omega$ is shunted the balance point will shift towards $3 \Omega$. So, new length $l^{\prime}=22.5+I=62.5$
So, $\frac{62.5}{37.5}=\frac{2}{3 x}(3+x)$
$\Rightarrow x=2 \Omega$
Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3$\Omega$. Cell 2 has emf of 6V and internal resistance of 6$\Omega$. An external resistor R of 4$\Omega$ is connected across A and B. The current flowing through R will be ____________ A.
Explanation:
$\mathrm{KCL}$ at $A$ gives
$\frac{6-V_{A}}{4}+\frac{0-V_{A}}{6}+\frac{18-V_{A}}{3}=0$
$V_{A}=10$
So current through $4 \Omega=\frac{10-6}{4}=1 \mathrm{~A}$
In the given circuit, the equivalent resistance between the terminal A and B is __________ $\Omega$.
Explanation:
Remove the resistors that have no current. Now the equivalent circuit becomes -
$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=3+(2 \| 2)+6 \\\\ & \mathrm{R}_{\mathrm{eq}}=3+1+6 \\\\ & \mathrm{R}_{\mathrm{eq}}=10 \Omega \end{aligned} $
If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is __________%.
Explanation:
Considering volume to be conserved
$ \begin{aligned} & \text { Vol. }=\ell_{0} A_{0}=\left(1.2 \ell_{0}\right) \mathrm{A} \\\\ & A_{\text {final }}=\frac{A_{0}}{1.2} \\\\ & R_{\text {in }}=\frac{\rho \ell_{0}}{A_{0}} \\\\ & R_{\text {final }}=\frac{\rho 1.2 \ell_{0}}{\frac{A_{0}}{1.2}}=\frac{\rho \ell_{0}}{A_{0}}(1.2)^{2} \end{aligned} $
$=\mathrm{R}_{\text {in }}(1.44)$
Hence increase $=44 \%$
A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is $n\times10^{-3}\Omega$. If the resistivity of the material is $\mathrm{2.4\times10^{-8}\Omega m}$. The value of $n$ is ___________.
Explanation:
$R=\rho \frac{l}{\pi\left(r_2^2-r_1^2\right)}$
where $r_2=$ outer radius
$ \begin{gathered} r_1=\text { inner radius } \\\\ \rho=\text { resistivity, } l=\text { length } \\\\ \therefore \rho=\frac{2.4 \times 10^{-8} \times 3.14}{\pi\left(\frac{8^2}{4}-\frac{4^2}{4}\right) \times 10^{-6}} \\\\ =\frac{2.4 \times 10^{-8} \times 4}{48 \times 10^{-6}}=2 \times 10^{-3} \Omega \end{gathered} $