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} $
The current I flowing through the given circuit will be __________A.
Explanation:
All $9 ~\Omega$ resistances are in parallel
${R_{eq}} = 3\,\Omega $
$I = {6 \over 3}A = 2A$
An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _______ W.
Explanation:
${P_{100}} = {{{V^2}} \over {{R_{100}}}} \Rightarrow {R_{100}} = {{{V^2}} \over {{P_{100}}}}$
${P_{60}} = {{{V^2}} \over {{R_{60}}}} \Rightarrow {R_{60}} = {{{V^2}} \over {{P_{60}}}}$
${P_{net}} = {{{V^2}} \over {{R_{60}} + {R_{100}}}} = {{{P_{60}}{P_{100}}} \over {{P_{60}} + {P_{100}}}} = {{60 \times 100} \over {160}} = 37.5$
This power developed is proportional to resistance.
So, $P{'_{60}} = {P_{net}} \times {{60} \over {160}} = 37.5 \times {{60} \over {160}} \simeq 14\,W$
As shown in the figure, a potentiometer wire of resistance $20 \,\Omega$ and length $300 \mathrm{~cm}$ is connected with resistance box (R.B.) and a standard cell of emf $4 \mathrm{~V}$. For a resistance '$R$' of resistance box introduced into the circuit, the null point for a cell of $20 \,\mathrm{mV}$ is found to be $60 \mathrm{~cm}$. The value of '$R$' is ___________ $\Omega .$
Explanation:
$l = 3m$, ${R_w} = 20\,\Omega $
${\varepsilon _0} = 4V$
${{4 \times 20} \over {20 + R}} \times {{60} \over {300}} = 20 \times {10^{ - 3}}$
${4 \over {20 + R}} = 5 \times {10^{ - 3}}$
$20 + R = 800$
$R = 780\,\Omega $
In the given figure of meter bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is $40 \mathrm{~cm}$. The balancing length, if the radius of the wire $\mathrm{AB}$ is doubled, will be ______________ $\mathrm{cm}$.
Explanation:
Even if the radius of wire is doubled, the balancing point would not change as ${x \over {l - x}} = {{{R_1}} \over {{R_2}}}$, which is not including a term of area.
In a meter bridge experiment, for measuring unknown resistance 'S', the null point is obtained at a distance $30 \mathrm{~cm}$ from the left side as shown at point D. If R is $5.6$ $\mathrm{k} \Omega$, then the value of unknown resistance 'S' will be __________ $\Omega$.
Explanation:
${R \over S} = {{70} \over {30}}$
$S = {3 \over 7} \times 5.6 \times {10^3} = 2.4 \times {10^3}\,\Omega $
$ = 2400\,\Omega $
A $1 \mathrm{~m}$ long copper wire carries a current of $1 \mathrm{~A}$. If the cross section of the wire is $2.0 \mathrm{~mm}^{2}$ and the resistivity of copper is $1.7 \times 10^{-8}\, \Omega \mathrm{m}$, the force experienced by moving electron in the wire is ____________ $\times 10^{-23} \mathrm{~N}$.
(charge on electorn $=1.6 \times 10^{-19} \,\mathrm{C}$)
Explanation:
$I = ne{v_d}A$
$J = {E \over \rho }$
$F = eE = {{1.7 \times 1.6 \times {{10}^{ - 19}} \times {{10}^{ - 8}}} \over {2 \times {{10}^{ - 6}}}}$
$ = 136 \times {10^{ - 23}}$ N
A potentiometer wire of length $300 \mathrm{~cm}$ is connected in series with a resistance 780 $\Omega$ and a standard cell of emf $4 \mathrm{V}$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $20\, \mathrm{mV}$ is found to be $60 \mathrm{~cm}$. The resistance of the potentiometer wire is ____________ $\Omega$.
Explanation:
$l = 300$ cm
$\varepsilon = Kx$
$20 \times {10^{ - 3}} = \left( {{{4 \times R} \over {780 + R}} \times {1 \over {300}}} \right)60$
$R = 20$
Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is $40 \mathrm{~cm}$. Now an unknown resistance $x$ is connected in series with $\mathrm{P}$ and new balancing length is found to be $80 \mathrm{~cm}$ measured from the same end. Then the value of $x$ will be ____________ $\Omega$.
Explanation:
${P \over {40}} = {Q \over {60}}$ ...... (1)
${{P + x} \over {80}} = {Q \over {20}}$ ..... (2)
${P \over {P + x}} \times {{80} \over {40}} = {{20} \over {60}}$
${4 \over {4 + x}} \times 2 = {1 \over 3}$
$24 = 4 + x$
$x = 20$
In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ___________ cm.
Explanation:
$E \propto I$
${{1.2} \over {1.8}} = {{36} \over {I'}}$
$I' = {3 \over 2} \times 36 = 54$ cm
$\Delta I = I' - I = 54 - 36 = 18$ cm
In the given figure, the value of Vo will be _____________ V.
Explanation:
Using Kirchhoff's junction rule.
${{2 - {V_0}} \over 1} + {{4 - {V_0}} \over 1} + {{6 - {V_0}} \over 1} = 0$
$12 - 3{V_0} = 0$
${V_0} = 4\,V$
Eight copper wire of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$. If a single copper wire of length $2 l$ have the same resistance $(R)$ then its diameter will be ____________ d.
Explanation:
$RAB = R$
$R = {1 \over 8}$ (Resistance of one wire)
$ = {1 \over 8}\rho {l \over {\pi {{{d^2}} \over 4}}} = {{\rho l} \over {2\pi {d^2}}}$
Resistance of copper wire of length $2l$ and diameter $x = R$
$\rho {{2l} \over {\pi {{{x^2}} \over 4}}} = R$
${{8\rho l} \over {\pi {x^2}}} = {{\rho l} \over {2\pi {d^2}}}$
$16{d^2} = {x^2}$
$x = 4d$
The circuit diagram of potentiometer used to measure the internal resistance of a cell (E) is shown in figure. The key 'K' is kept closed so as to send constant current through potentiometer wire. When key 'K1' is kept open the null point is found to be at 120 cm on the potentiometer wire. When the key 'K1' is closed the null point is shifted at 80 cm at the potentiometer wire. The internal resistance of the given cell is _____________ $\Omega$.
Explanation:
Shortcut Method :
Internal Resistance of Unknown Battery
$r=\left(\frac{\ell_1-\ell_2}{\ell_2}\right) \mathrm{R}$
Where l1 means balanced length when key K1 is open
Where l2 means balanced length when key K1 is closed
Here l1 = 120 cm and l2 = 80 cm and R = 4 $\Omega $
$ \therefore $ $r = \left( {{{120 - 80} \over {80}}} \right) \times 4$ = 2 $\Omega $
Normal Method :
Original Potentiometer is $\to$
Here, AB is called potentiometer wire. Let length of AB = l and resistance = R
$\therefore$ (1) Current in the circuit (i) = ${{4} \over {R + {R_h}}}$
(2) Potential difference between A and B is
= VA $-$ VB = $\Delta$VAB = i $\times$ R
(3) ${{\Delta {V_{AB}}} \over {AB}} = {{Potential\,difference} \over {Length}}$ = Potential Gradient
$ \Rightarrow {{\Delta {V_{AB}}} \over l} = {{iR} \over l} = {4 \over {R + {R_h}}} \times {R \over l}$ Volt/m
Now, a circuit added to potentiometer wire.
When K1 is open then circuit look like the following.
When null point is found at C then no current will flow through the circuit added to the potentiometer wire.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 120 = {{4R} \over {(R + {R_h})l}} \times 120$ ...... (1)
(2) From external circuit :
VA $-$ 1.5 $-$ 0 $\times$ r = VC
$\Rightarrow$ VA $-$ VC = 1.5V = $\Delta$VAC ...... (2)
Comparing equation (1) and (2), we get
${{4R} \over {(R + {R_h})l}} \times 120 = 1.5$ ..... (3)
When the key K1 is closed then circuit look like this $\to$
When null point is found at C then no current will flow through the circuit added to the potentiometer wire but i1 current will flow through circuit created using 1.5V with internal resistance r and 4$\Omega$ resistance.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 80 = {{4R} \over {(R + {R_h})l}} \times 80$ ....... (4)
(2) From external circuit :
${i_1} = {{1.5} \over {4 + r}}$
${V_A} - 1.5 + {i_1}r = {V_C}$
$ \Rightarrow {V_A} - {V_C} = 1.5 - {i_1}r = 1.5 - {{1.5r} \over {4 + r}}$ ..... (5)
Or we can use this also,
${V_A} - {i_1} \times 4 = {V_C}$
$ \Rightarrow {V_A} - {V_C} = {i_1} \times 4 = {{1.5} \over {4 + r}} \times 4$
From equation 4 and 5, we get
${{4R} \over {(R + {R_h})l}} \times 80 = 1.5 - {{1.5r} \over {4 + r}}$
$ \Rightarrow {{1.5} \over {120}} \times 80 = 1.5 - {{1.5r} \over {4 + r}}$ [From equation 3]
$ \Rightarrow 1.5 \times {2 \over 3} = 1.5 - {{1.5r} \over {4 + r}}$
$ \Rightarrow {{1.5r} \over {4 + r}} = 1.5 - 1.5 \times {2 \over 3}$
$ \Rightarrow {{1.5r} \over {4 + r}} = 1.5 \times {1 \over 3}$
$ \Rightarrow 3r = 4 + r$
$ \Rightarrow 2r = 4$
$ \Rightarrow r = 2\,\Omega $
Other Method :
Now this can be converted to single battery and single resistance using combination of battery rule,
We know,
${E_{eq}} = {{{E_1}{r_2} + {E_2}{r_1}} \over {{r_1} + {r_2}}}$
and ${r_{eq}} = {{{r_1}{r_2}} \over {{r_1} + {r_2}}}$
$\therefore$ Here, ${E_{eq}} = {{1.5 \times 4 + 0 \times r} \over {4 + r}} = {6 \over {4 + r}}$
and ${r_{eq}} = {{4r} \over {4 + r}}$
$\therefore$ Circuit becomes
When null point is found at C then no current will flow through the circuit added to the potentiometer wire.
Potential difference between A and C is
(1) From potentiometer wire :
${V_A} - {V_C} = \Delta {V_{AC}} = {{{V_{AB}}} \over l} \times 80 = {{4R} \over {(R + {R_h})l}} \times 80$ .... (4)
(2) From external circuit :
${V_A} - {6 \over {4 + r}} - 0 \times {{4r} \over {4 + r}} = {V_C}$
$ \Rightarrow {V_A} - {V_C} = {6 \over {4 + r}}$ ...... (5)
From equation 4 and 5, we get
${{4R} \over {(R + {R_h})l}} \times 80 = {6 \over {4 + r}}$
$ \Rightarrow {{1.5} \over {120}} \times 80 = {6 \over {4 + r}}$ [From equation 3]
$ \Rightarrow 1.5 \times {2 \over 3} = {6 \over {4 + r}}$
$ \Rightarrow 4 + r = 6$
$ \Rightarrow r = 2\,\Omega $
Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $\Omega$ is used to measure the potential difference across 500 $\Omega$ resistor, the reading of the voltmeter will be ___________ V.
Explanation:
New ${R_{eff}} = {{2000 \times 500} \over {2500}} + 600\,\Omega = 1000\,\Omega $
$\Rightarrow$ Reading of voltmeter $ = {{400} \over {1000}} \times 20 = 8$ volts
The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x $\times$ 10$-$3 $\Omega$ cm. The value of x is ____________. [Take $\pi$ = 3.14]
Explanation:
Resistance $ = \tan 45^\circ = 1\,\Omega $
$ \Rightarrow 1 = {{pI} \over A}$
$ \Rightarrow p = {{\pi {{(1.2\,cm)}^2}} \over {31.4\,cm}} = 1.44 \times {10^{ - 1}}$ $\Omega$ cm
$ \Rightarrow x = 144$
For the network shown below, the value of VB $-$ VA is ____________ V.
Explanation:
${V_B} - {V_A} = i \times 2$
$ = {{15} \over {1 + 2}} \times 2$
$ \Rightarrow {V_B} - {V_A} = 10$ volts
All resistances in figure are 1 $\Omega$ each. The value of current 'I' is ${a \over 5}$ A. The value of a is _________.
Explanation:
Let the current is i
Using Kirchhoff's law
$iR + {i \over 2}R + {i \over 4}R + {i \over 8}R = 3$
$i = {{3 \times 8} \over {15}} = {8 \over 5}A$
So $a = 8$
A meter bridge setup is shown in the figure. It is used to determine an unknown resistance R using a given resistor of 15 $\Omega$. The galvanometer (G) shows null deflection when tapping key is at 43 cm mark from end A. If the end correction for end A is 2 cm, then the determined value of R will be ____________ $\Omega$.
Explanation:
${{43 + 2} \over {15}} = {{57} \over R}$
$R = {{57 \times 15} \over {45}} = 19\,\Omega $
Current measured by the ammeter (A) in the reported circuit when no current flows through 10 $\Omega$ resistance, will be ________________ A.
Explanation:
For ${I_{10}} = 0$
${R \over 3} = {4 \over 6}$
$ \Rightarrow R = 2\,\Omega $
$ \Rightarrow {I_A} = {{36 \times (6 + 9)} \over {6 \times 9}}$
$ = {{36 \times 15} \over {6 \times 9}} = 10$ A
The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 $\times$ 106 A/m2. The current through the outer portion of the wire between radial distances ${r \over 2}$ and r is x$\pi$ A; where x is __________.
Explanation:
$i = A \times j$
$ = \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j$
$ = {{3\pi {R^2}} \over 4} \times j$
$ = {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 1.0 \times {10^6}$
$ = 12\,\pi $
In the given circuit 'a' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be $\sqrt {{x \over 2}} $. The value of x is __________.
Explanation:
${R_{net}} = {{ma} \over 3} + {a \over {2m}}$
$ = a\left[ {{m \over 3} + {1 \over {2m}} - {2 \over {\sqrt 6 }} + {2 \over {\sqrt 6 }}} \right]$
$ = a\left[ {{{\left( {\sqrt {{m \over 3}} - {1 \over {\sqrt {2m} }}} \right)}^2} + \sqrt {{2 \over 3}} } \right]$
This will be minimum when
$\sqrt {{m \over 3}} = {1 \over {\sqrt {2m} }}$
or $m = \sqrt {{3 \over 2}} $
so $x = 3$
A cell, shunted by a 8 $\Omega$ resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 $\Omega$ resistance. The value of internal resistance of the cell will be ____________ $\Omega$.
Explanation:
${{{\varepsilon _1}8} \over {{r_1} + 8}} =3c$
${{{\varepsilon _1}4} \over {{r_1} + 4}} =2c$
$ \Rightarrow {{2({r_1} + 4)} \over {{r_1} + 8}} = {3 \over 2}$
$ \Rightarrow {r_1} = 8\,\Omega $
The current density in a cylindrical wire of radius 4 mm is 4 $\times$ 106 Am$-$2. The current through the outer portion of the wire between radial distances ${R \over 2}$ and R is ____________ $\pi$ A.
Explanation:
$i = A \times j$
$ = \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j$
$ = {{3\pi {R^2}} \over 4} \times j$
$ = {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 4 \times {10^6}$
$ = 48\pi $
The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ____________ %.
Explanation:
Volume is constant so on length doubled
Area is halved so
$R = \rho {l \over A}$ and $R' = \rho {{2l} \over {{A \over 2}}} = 4\rho {l \over A} = 4R$
So percentage increase will be
$R\% = {{4R - R} \over R} \times 100 = 300\% $
A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ____________ J.
Explanation:
$300 = {I^2}R \times 15$
$ \Rightarrow R = 5\,\Omega $
Now $I_2^2R{t_2}$
$ = 9 \times 5 \times 10$
$ = 450\,J$
The total current supplied to the circuit as shown in figure by the 5 V battery is ____________ A.
Explanation:
The equivalent circuit is
$ \Rightarrow I = {5 \over {2.5}} = 2\,A$
A potentiometer wire of length 10 m and resistance 20 $\Omega$ is connected in series with a 25 V battery and an external resistance 30 $\Omega$. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is ${x \over {10}}$. The value of x is __________.
Explanation:
$\therefore$ $E = I \times \left( {{{20} \over 4}} \right) = {{25} \over {(30 + 20)}} \times \left( {{{20} \over 4}} \right)$
$ = {1 \over 2} \times 5 = 2.5$ volts
$ = {{25} \over {10}}$ volts
In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.
Explanation:
At balancing point, we know that emf is proportional to the balancing length. i.e.,
emf $\propto$ balancing length
Now, let the emf's be 3$\varepsilon $ and 2$\varepsilon $.
$\Rightarrow$ 3$\varepsilon $ = k(75) ..... (1)
and 2$\varepsilon $ = k(l) ....... (2)
$\Rightarrow$ l = 50 cm
$\Rightarrow$ Difference is (75 $-$ 50) cm = 25 cm.
Explanation:
The potential difference of the uniform wire, V = 240 V
The resistance of the uniform wire, R1 = 36 $\Omega$
The power dissipation in the first case,
${P_1} = {{{V^2}} \over {{R_1}}} = {{{{(240)}^2}} \over {36}}$
For Case II,
The resistance of each half, ${R_2} = {{{R_1}} \over 2} = {{36} \over 2} = 18\Omega $
${P_2} = {{{V^2}} \over {{R_2}}} + {{{V^2}} \over {{R_2}}} = {{{{(240)}^2}} \over {18}} + {{{{(240)}^2}} \over {18}} = {{{{(240)}^2}} \over 9}$
Thus, the ratio of the total power dissipation in the first case to the second case
${{{P_1}} \over {{P_2}}} = {{{{(240)}^2}/36} \over {{{(240)}^2}/9}} \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over 4}$
Comparing with, ${{{P_1}} \over {{P_2}}} = {1 \over x}$
The value of the x = 4.
Explanation:
192 = 16 (R) (1)
R = 12$\Omega$
E1 = (8)2 (12) (5)
= 3840 J
Explanation:
Req = 3$\Omega$