Current Electricity

Series Arrangement

For cells in series, the total electromotive force (EMF) is doubled, and the total internal resistance is also doubled. The current $I_1$ in this setup is given by:

$ I_1 = \frac{2E}{2 + 2r} $

where $E$ is the electromotive force of each cell and $r$ is the internal resistance of each cell.

Parallel Arrangement

For cells in parallel, the EMF remains the same, but the internal resistance is halved. The current $I_2$ in this setup is given by:

$ I_2 = \frac{E}{2 + \frac{r}{2}} = \frac{2E}{4 + r} $

Equating the Currents

Since the problem states $I_1 = I_2$, we set the equations equal:

$ \frac{2E}{4 + r} - \frac{2E}{2 + 2r} = 0 $

Solving this equation:

$ \frac{2E(2 + 2r) - 2E(4 + r)}{(4 + r)(2 + 2r)} = 0 $

Simplifying further:

$ 2E(2 + 2r) - 2E(4 + r) = 0 $

$ 4E + 4Er - 8E - 2Er = 0 $

$ 2Er - 4E = 0 $

Factoring out common terms:

$ Er - 2E = 0 $

Divide by $E$ (assuming $E \neq 0$):

$ r - 2 = 0 $

Hence, solving this gives:

$ r = 2 $

Thus, the internal resistance of each cell is $2 \, \Omega$.

Given that the charge $ q $ passing through a $10 \, \Omega$ resistor as a function of time $ t $ is expressed by the equation $ q = 3t^2 - 2t + 6 $, we can determine the potential difference across the resistor.

To find the current $ I $ as a function of time, we differentiate the charge function with respect to time:

$ I = \frac{dq}{dt} = \frac{d}{dt}(3t^2 - 2t + 6) $

Solving for the derivative:

$ I = 6t - 2 $

Next, we calculate the current at $ t = 5 \, \text{s} $:

$ I = 6(5) - 2 = 30 - 2 = 28 \, \text{A} $

The voltage $ V $ across the resistor, using Ohm's Law, is given by:

$ V = IR $

With the resistance $ R = 10 \, \Omega $, the potential difference is:

$ V = 28 \times 10 = 280 \, \text{V} $

Thus, the potential difference across the resistor at $ t = 5 \, \text{s} $ is $ 280 \, \text{V} $.

Given the following values:

$E_1 = 1.2 \, \text{V}$ with an internal resistance $r_1 = 2 \, \Omega$

$E_2 = 1.5 \, \text{V}$ with an internal resistance $r_2 = 1 \, \Omega$

To find the equivalent emf of the cells connected in parallel with like poles together, we first calculate the equivalent resistance as follows:

$ r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3} \, \Omega $

Next, we calculate the equivalent emf using the formula:

$ \text{Emf (equivalent)} = \left[\frac{E_1}{r_1} + \frac{E_2}{r_2}\right] \times r_{\text{eq}} $

Substituting the given values, we have:

$ \begin{aligned} \text{Emf (equivalent)} & = \left[\frac{1.2}{2} + \frac{1.5}{1}\right] \times \frac{2}{3} \\ & = \left[0.6 + 1.5\right] \times \frac{2}{3} \\ & = 2.1 \times \frac{2}{3} \\ & = 1.4 \, \text{V} \end{aligned} $

Thus, the equivalent emf of the combination of the two cells is $1.4 \, \text{V}$.

Given, Voltage across battery $=V$

Each resistor have resistance $=2 R$

$ i_{F C}=\text { ? } $

According to figure, it is a balanced Wheatstone bridge.

$ \text { } \begin{aligned}So, R_1 & =R_{F C}+R_{C E}=2 R+2 R=4 R \\ R_2 & =R_{F D}+R_{D E}=2 R+2 R=4 R \\ R_{\text {eq }} & =\frac{R_1 \times R_2}{R_1+R_2}=\frac{4 R \times 4 R}{4 R+4 R}=\frac{16 R R}{8 R}=2 R \end{aligned} $

We know that, voltage will be same in parallel combination. So, $R_1$ will have voltage $V$.

Current will be same in series combination. So $R_{F C}$ and $R_{C E}$ will have current same as $R_1$

So, $i_{F C}=\frac{V}{R_1} \Rightarrow i_{F C}=\frac{V}{4 R}$

Given the specifications for the lamp:

Voltage, $ V = 240 \, \text{V} $

Power, $ P = 60 \, \text{W} $

The problem states that when the lamp is in use, the resistance of the filament (hot resistance, $ R_{\text{Hot}} $) is 20 times that of when it is not in use (cold resistance, $ R_{\text{Cold}} $).

Calculating Hot Resistance

Using the power formula from Joule's law of heating:

$ P = \frac{V^2}{R} $

Solve for $ R_{\text{Hot}} $:

$ R_{\text{Hot}} = \frac{V^2}{P} = \frac{240 \times 240}{60} = 960 \, \Omega $

Relating Hot and Cold Resistance

According to the problem, the hot resistance is 20 times the cold resistance:

$ R_{\text{Hot}} = 20 \times R_{\text{Cold}} $

Solving for $ R_{\text{Cold}} $:

$ R_{\text{Cold}} = \frac{R_{\text{Hot}}}{20} = \frac{960}{20} = 48 \, \Omega $

Thus, the resistance of the lamp when not in use is $ 48 \, \Omega $.

Since, wire is bent into form of a square so each side will have resistance,

$ r=\frac{R}{4} $

Here, $A D, D C$ and $C B$ are in series and $A B$ will be parallel,

$ \therefore R_{\text {eq }}=\frac{3 R}{4} \| \frac{R}{4}=\frac{\frac{3 R}{4} \times \frac{R}{4}}{\frac{3 R}{4}+\frac{R}{4}}=\frac{\frac{3 R^2}{16}}{R}=\frac{3 R}{16} $

Current in the circuit, $I=\frac{V}{R_{\text {eq }}}=\frac{12}{\frac{3 R}{16}}=\frac{4}{4}$

So, potential difference between end $A$ and $B$

$ =\frac{V}{R / 4}=\frac{48}{R} $

$\therefore$ Current in the branch

$ A D C=\frac{64}{R}-\frac{48}{R}=\frac{16}{R} $

$\therefore$ Potential dropp across any diagnal $A C$ or $B D$ )

$ =\frac{16}{R} \times \frac{R}{2}=8 \mathrm{~V} $

Applying KVL in the loop considering current in anti-clockwise direction.

$ \begin{aligned} -36+2 i+12+3 i & =0 \\ 5 i & =24 \\ i & =\frac{24}{5} \mathrm{~A} \end{aligned} $

$ \begin{aligned} & \begin{aligned} \text { Now; } V_B-V_A & =12+3 \times i \\ 24-V_A & =12+3 \times \frac{24}{5} \\ \Rightarrow \quad V_A & =24-12-\frac{72}{5}=12-\frac{72}{5} \\ & =-\frac{12}{5}=-2.4 \mathrm{~V} \end{aligned} \end{aligned} $

Given the block dimensions of $1 \, \mathrm{cm}$, $2 \, \mathrm{cm}$, and $3 \, \mathrm{cm}$, we need to determine the ratio of the maximum to minimum resistance between any pair of opposite faces of the block.

The resistance $R$ of a conductor is given by the formula:

$ R = \rho \frac{l}{A} $

where:

$\rho$ is the resistivity of the material,

$A$ is the area of cross-section,

$l$ is the length.

Calculating Maximum Resistance:

The resistance is maximum when:

$l = 3 \, \mathrm{cm} = 3 \times 10^{-2} \, \mathrm{m}$

$A = 1 \times 10^{-2} \times 2 \times 10^{-2} = 2 \times 10^{-4} \, \mathrm{m}^2$

Thus, the maximum resistance $R_{\max}$ is:

$ R_{\max} = \rho \times \frac{3 \times 10^{-2}}{2 \times 10^{-4}} = \frac{3}{2} \times 10^2 \rho $

Calculating Minimum Resistance:

The resistance is minimum when:

$l = 1 \, \mathrm{cm} = 10^{-2} \, \mathrm{m}$

$A = 2 \times 10^{-2} \times 3 \times 10^{-2} = 6 \times 10^{-4} \, \mathrm{m}^2$

Thus, the minimum resistance $R_{\min}$ is:

$ R_{\min} = \rho \times \frac{10^{-2}}{6 \times 10^{-4}} = \frac{10^2 \rho}{6} $

Calculating the Ratio:

The ratio of the maximum resistance to the minimum resistance is:

$ \frac{R_{\max}}{R_{\min}} = \frac{\frac{3 \times 10^2 \rho}{2}}{\frac{10^2 \rho}{6}} = \frac{3}{2} \times 6 = \frac{9}{1} $

Therefore, the ratio is $9:1$.

Using Kirchhoff's current law at point $P$,

$ I_1+I_2=6 $

Using Krichhoff's voltage law to the closed circuit $P Q R P$,

$ \begin{array}{rr} \Rightarrow & -2 I_1-2 I_1+2 I_2=0 \\ \Rightarrow & -4 I_1+2 I_2=0 \\ \Rightarrow & 2 I_1-I_2=0 \end{array} $

On adding Eqs. (i) and (ii), we get

$ \begin{array}{cc} \Rightarrow & I_1+I_2+2 I_1-I_2=6+0 \\ \Rightarrow & 3 I_1=6 \\ \Rightarrow & I_1=2 \mathrm{~A} \end{array} $

Substitute the value of $I_1$ in Eq. (i)

$ \Rightarrow \quad I_2=6-I_1=6-2=4 \mathrm{~A} $

Given, Galvanometer resistance, $G=99 \Omega$

Let $I$ be the current passes through main circuit.

$ \therefore \quad I_g=\left(\frac{10}{100}\right) I=0.1 I $

Using the formula for shunt resistance,

$ S=\frac{I_g G}{\left(I-I_g\right)} $

$ \begin{aligned} & \Rightarrow \quad S=\frac{0.1 I \times 99}{(I-0.1 I)}=\frac{0.1 \times 99}{0.9} \\ & \therefore \quad S=11 \Omega \end{aligned} $

In a steady current flowing through a conductor, the quantity that remains constant is the electric current. Here's why:

For any section of the conductor, the electric current $I$ is defined as the net charge per unit time passing through that section. In a steady state (no accumulation of charge), charge conservation requires that $I$ is constant along the length of the conductor.

The current density $J$ is given by

$J = \frac{I}{A},$

where $A$ is the local cross-sectional area. If the conductor has a non-uniform cross-section, the area $A$ changes, causing $J$ to vary.

Similarly, the drift velocity $v_d$ of the charge carriers is related to the current by

$I = n e v_d A,$

where $n$ is the carrier density and $e$ is the elementary charge. Again, if $A$ varies, $v_d$ must adjust accordingly to keep $I$ constant.

Thus, it is only the electric current $I$ that remains unchanged along a conductor with non-uniform cross-section.

The correct answer is:

Option D

electric current

The circuit diagram given below represents balanced Wheatstone bridge.

$\therefore \quad R_{A C}=\frac{2 R \times 2 R}{2 R+2 R}=\frac{4 R^2}{4 R}=R$

Since, potential barrier in Si diode is less $(0.7 \mathrm{~V})$ than green, and both are forward biased. Therefore, for conduction, minimum potential difference across the junction diode $=0.7 \mathrm{~V}$.

$ \therefore \quad V_0=12-0.7=11.3 \mathrm{~V} $

To find the resistivity of the potentiometer wire, we are provided with the following information:

Area of cross-section (A): $ 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 $

Current (i): $ 1 \, \text{A} $

Potential Gradient $\left(\frac{V}{l}\right)$: $ 7.5 \, \text{V/m} $

We need to calculate the resistivity ($\rho$) of the wire.

Calculation:

From Ohm's law, we know the following relationship:

$ V = i \cdot R $

The resistance $R$ for a wire can be expressed in terms of its resistivity:

$ R = \rho \frac{l}{A} $

Thus, substituting $R$ in the equation for $V$:

$ V = i \cdot \frac{\rho l}{A} $

Rearranging this equation to solve for resistivity ($\rho$):

$ \rho = \frac{V}{l} \cdot \frac{A}{i} $

Substitute the known values:

$ \rho = \left(7.5 \, \text{V/m}\right) \cdot \left(\frac{4 \times 10^{-4} \, \text{m}^2}{1 \, \text{A}}\right) $

Calculate $\rho$:

$ \rho = \frac{7.5 \times 4 \times 10^{-4}}{1} $

$ \rho = 30 \times 10^{-4} $

$ \rho = 3 \times 10^{-3} \, \Omega\text{-m} $

Therefore, the resistivity of the potentiometer wire is $3 \times 10^{-3} \, \Omega\text{-m}$.

Drift velocity is the average velocity acquired by charged particles, such as electrons, within a material when influenced by an electric field.

The magnitude of drift velocity per unit electric field is defined as mobility. This relationship is given by:

$ m = \frac{|v_d|}{E} \Rightarrow |v_d| = m \cdot E $

From this, we understand that:

$ v_d \propto E $

Thus, the drift velocity is directly proportional to the electric field $ E $. Therefore, the correct relationship is represented by $ v \propto E $.

To find the average value of the current from $ t = 0 $ to $ t = 2 $ seconds, we analyze the current-time ($ i-t $) graph.

Calculate the Area under the Graph:

The graph consists of two triangular sections.

First Triangle: From $ t = 0 $ to $ t = 1 $ second, with a height of 10 A.

$ \text{Area of first triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 10 = 5 \text{ A} $

Second Triangle: From $ t = 1 $ to $ t = 2 $ seconds, with a height of 10 A.

$ \text{Area of second triangle} = \frac{1}{2} \times (2-1) \times 10 = 5 \text{ A} $

Total Area:

Total area under the graph from $ t = 0 $ to $ t = 2 $ seconds is:

$ 5 \text{ A} + 5 \text{ A} = 10 \text{ A} $

Calculate the Average Current:

The average current $(i_{\mathrm{av}})$ is given by:

$ i_{\mathrm{av}} = \frac{\text{Total Area}}{\text{Time Interval}} = \frac{10}{2} = 5 \text{ A} $

Thus, the average current over the interval from $ t = 0 $ to $ t = 2 $ seconds is $ 5 \, \text{A} $.

The given circuit can be redrawn as,

where, equivalent emf of the combination,

$ \begin{aligned} E_{\mathrm{eq}} & =\frac{E_{1} R_{1}+E_{2} R_{2}}{R_{1}+R_{2}} \\ & =\frac{2 \times 4+2 \times 4}{4+4}=2 \mathrm{~V} \end{aligned} $

and $ R_{e q}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{4 \times 4}{4+4}=\frac{16}{8}=2 \Omega $

Net emf of the circuit,

$ \begin{aligned} E^{\prime} & =E_{2}+E_{\mathrm{eq}} \\ & =2+2 \\ & =4 \mathrm{~V} \end{aligned} $

$ \therefore $ Current flowing through branch $ A B $,

$ I=\frac{E^{\prime}}{R_{\mathrm{eq}}}=\frac{4}{2}=2 \mathrm{~A} $

Clearly from figure, it is clear that direction of current is from point $ A $ to $ B $ clockwise is loop.

The equivalent Wheatstone's bridge network of the given circuit is shown in figure. Here,

The points $ B $ and $ D $ are at the same potential as the bridge is balanced. So, the $ 3 \Omega $ resistance in $ B D $ arm is ineffective and can be omitted from the circuit.

Hence, equivalent resistance remains same when $ 3 \Omega $ resistor is removed.

When a conducting wire is stretched due to a deforming force, its diameter decreases to 40% of its original value. We need to determine the percentage change in its resistance.

Conceptual Framework:

Volume Conservation:

When the wire is stretched, its volume ($V$) remains constant. Therefore, the relationship between volume, cross-sectional area ($A$), and length ($l$) is given by:

$ V = A \times l $

Solving for $l$, we have:

$ l = \frac{V}{A} $

Resistance Formula:

The resistance ($R$) of a wire is determined by its resistivity ($\rho$), length, and cross-sectional area:

$ R = \rho \cdot \frac{l}{A} = \frac{\rho V}{A^2} $

Given the relation between area and diameter ($D$), we express $A$ in terms of $D$:

$ A = \frac{\pi D^2}{4} $

Thus, resistance can be reformulated as:

$ R = \frac{\rho V}{A^2} = \frac{16 \rho V}{\pi^2 D^4} $

Change in Resistance:

The percentage change in resistance can be calculated using the change in diameter:

$ \frac{\Delta R}{R} = -4 \frac{\Delta D}{D} $

Given the diameter decreases to 40% of its original value, the change in diameter is $\Delta D = -0.6D$.

Substituting this into our percentage change formula, we get:

$ \frac{\Delta R}{R} = -4(-0.6) = 4 \times 0.6 = 1.6\% $

Therefore, the percentage change in resistance as a result of the stretching is 1.6%.

Assertion A is incorrect because the preferred voltmeter for measuring the potential difference across a resistance of 600 ohm is actually the voltmeter with a higher resistance, not a lower resistance. The reason for this is that when a voltmeter is connected in parallel with the resistance being measured, it will draw current away from the resistance, reducing the potential difference across it. A higher resistance voltmeter will draw less current and therefore have a smaller effect on the potential difference being measured.

In this case, the voltmeter with resistance 4000 ohm would be preferred because it would draw less current than the voltmeter with resistance 1000 ohm, leading to a more accurate measurement of the potential difference across the resistance of 600 ohm.

Therefore, assertion A is incorrect because it states that the voltmeter with resistance 1000 ohm is preferred, when in reality the voltmeter with resistance 4000 ohm is preferred.

Reason R states that a voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance. This is correct because Ohm's law states that the current through a resistor is proportional to the voltage across it and inversely proportional to the resistance. Thus, a voltmeter with higher resistance will draw less current, making it a better choice for measuring the potential difference across another resistance.

Equivalent circuit can be drawn as

$\therefore$ $\frac{1}{R_{AB}}=\frac{7}{12}+\frac{1}{4}+\frac{2}{3}$

$R_{AB}=\frac{2}{3}~\Omega$

$Q(t) = \alpha t - \beta {t^2} + \gamma {t^3}$

$i(t) = \alpha - 2\beta t + 3\gamma {t^2}$

${{di} \over {dt}} = - 2\beta + 6\gamma t = 0$ (for max/min of i)

at $t = {\beta \over {3r}}$ (i is minimum as i is an upward parabola)

$i\left( {{\beta \over {3\gamma }}} \right) = \alpha - 2\beta \left( {{\beta \over {3\gamma }}} \right) + {{3\gamma {\beta ^2}} \over {9{\gamma ^2}}}$

$ = \alpha {{ - {\beta ^2}} \over {3\gamma }}$

A potentiometer is an electrical instrument used to measure the electromotive force (EMF) of a cell.

The sensitivity of a potentiometer is defined as the change in potential difference per unit length of the wire.

It is directly proportional to the length of the potentiometer wire (Option A), because a longer wire has a larger potential difference that can be measured.

It is inversely proportional to the potential gradient of the wire (Option C), because a smaller potential gradient results in a smaller change in potential difference per unit length.

Therefore, the correct answer is (C) A and C only.

$ \mathrm{R}_{\text {initial }}=\frac{\rho \ell}{A}=5 \Omega $

$\because$ Volume of wire is constant in stretching

$ \begin{aligned} & \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\\\ & \mathrm{A}_{\mathrm{i}} \ell_{\mathrm{i}}=\mathrm{A}_{\mathrm{f}} \ell_{\mathrm{f}} \\\\ & \mathrm{A} \ell=\mathrm{A}^{\prime}(5 \ell) \\\\ & \mathrm{A}^{\prime}=\frac{\mathrm{A}}{5} \\\\ & \mathrm{R}_{\mathrm{f}}=\frac{\rho \ell_{\mathrm{f}}}{\mathrm{A}_{\mathrm{f}}}=\frac{\rho(5 \ell)}{\left(\frac{\mathrm{A}}{5}\right)} \\\\ & =25\left(\frac{\rho \ell}{\mathrm{A}}\right) \\\\ & =25 \times 5=125 \Omega \end{aligned} $