Vardaan Learning Institute

Ans: Average current is the total charge flowing over a macroscopic time interval: $I_{avg} = \frac{\Delta q}{\Delta t}$. Instantaneous current is the rate of flow of charge at a specific instant of time, found using limits: $I = \lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t} = \frac{dq}{dt}$.

Ans: Current $I = 1.5\text{ A}$, which means $1.5\text{ C}$ of charge flows per second ($t=1\text{s}$).
$q = ne \implies n = \frac{q}{e} = \frac{1.5}{1.6 \times 10^{-19}} = 0.9375 \times 10^{19} = 9.375 \times 10^{18}\text{ electrons}$.

Ans: Instantaneous current $I = \frac{dq}{dt} = \frac{d}{dt}(\alpha t - \frac{1}{2}\beta t^2) = \alpha - \beta t$.
Initial current (at $t=0$): $I_0 = \alpha - \beta(0) = \alpha$.
Time when $I=0$: $0 = \alpha - \beta t \implies t = \frac{\alpha}{\beta}$.

Ans: Current density ($\vec{J}$) at a point is a vector whose magnitude is the electric current passing per unit area normal to the direction of current. $J = I/A$. It is uniform if the current is distributed evenly across a uniform, constant cross-sectional area.

Ans: Time period $T = \frac{2\pi r}{v}$. Equivalent current $I = \frac{e}{T} = \frac{e v}{2\pi r}$.
$I = \frac{(1.6 \times 10^{-19}) \times (2.2 \times 10^6)}{2 \times 3.14 \times (0.5 \times 10^{-10})} = \frac{3.52 \times 10^{-13}}{3.14 \times 10^{-10}} \approx 1.12 \times 10^{-3}\text{ A} = 1.12\text{ mA}$.

Ans: No. By conservation of charge in a steady state, the total current $I$ remains constant along the wire. Since Current Density $J = I/A$, as the cross-sectional area $A$ varies, the current density $J$ must vary inversely. ($J \propto 1/A$).

Ans: In a V-I graph (V on Y-axis, I on X-axis), the slope is $\frac{\Delta V}{\Delta I}$, which represents the Resistance ($R$) of the conductor. If axes are swapped (I on Y-axis, V on X-axis), the slope is $\frac{\Delta I}{\Delta V}$, which represents the inverse of resistance ($1/R$), called Conductance.

Ans: When stretched, volume remains constant ($A \times l = A' \times l'$). If $l' = 3l$, then $A' = A/3$.
Original $R = \rho \frac{l}{A} = 10\text{ }\Omega$.
New $R' = \rho \frac{l'}{A'} = \rho \frac{3l}{A/3} = 9 \left(\rho \frac{l}{A}\right) = 9 \times 10 = 90\text{ }\Omega$.

Ans: Start with $V = IR$. Substitute $V = El$ and $R = \rho\frac{l}{A}$.
$El = I (\rho \frac{l}{A}) \implies E = \rho (\frac{I}{A})$.
Since $J = I/A$, we get $E = \rho J$.
Since conductivity $\sigma = 1/\rho$, we get $E = \frac{J}{\sigma}$ or $J = \sigma E$. In vector notation, direction of $\vec{J}$ is same as $\vec{E}$, so $\vec{J} = \sigma\vec{E}$.

Ans: $R = \rho \frac{l}{\pi r^2}$. For B: $R_B = \rho \frac{l_B}{\pi r_B^2}$.
For A: $l_A = 2l_B$, $r_A = r_B/2$. Area $A_A = \pi (r_B/2)^2 = \pi r_B^2 / 4 = A_B / 4$.
$R_A = \rho \frac{2l_B}{A_B/4} = 8 \left(\rho \frac{l_B}{A_B}\right) = 8 R_B$. Ratio $R_A : R_B = 8:1$.

Ans: Resistivity ($\rho$) is an intrinsic property. It depends *only* on the nature of the material (electron density $n$) and its temperature (relaxation time $\tau$). It is independent of dimensions (length, area).

Ans: Area $A = \pi(d/2)^2 = \frac{\pi d^2}{4}$. Resistance $R \propto 1/A \implies R \propto 1/d^2$. If diameter $d$ is doubled ($2d$), the resistance becomes $1/4$th of its original value.

Ans: Consider a conductor of length $l$ and area $A$. Volume $V = Al$. Total number of free electrons $N = n \times Al$ (where $n$ is electron density). Total charge $q = Ne = neAl$.
Time taken by an electron to cross length $l$ is $t = l/v_d$.
Current $I = \frac{q}{t} = \frac{neAl}{l/v_d} = neAv_d$.

Ans: Relaxation time ($\tau$) is the average time between successive collisions. When temperature increases, thermal vibration of metal ions increases, causing electrons to collide more frequently. Thus, relaxation time $\tau$ decreases.

Ans: $I = neAv_d = neA(\frac{eE\tau}{m}) = \frac{ne^2A\tau}{m} E$.
Substitute $E = V/l$: $I = \left(\frac{ne^2A\tau}{ml}\right) V \implies V = \left(\frac{m}{ne^2\tau}\right) \frac{l}{A} I$.
This is $V = IR$, proving Ohm's Law where $R = \left(\frac{m}{ne^2\tau}\right) \frac{l}{A}$. Comparing with $R = \rho\frac{l}{A}$, resistivity $\rho = \frac{m}{ne^2\tau}$.

Ans: Only the total **Current ($I$)** remains constant (due to charge conservation). Current Density ($J = I/A$), Drift Speed ($v_d = I/neA$), and Electric Field ($E = J/\sigma$) all change as Area ($A$) changes.

Ans: $I = neAv_d \implies v_d = \frac{I}{neA}$.
$v_d = \frac{1.5}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1.0 \times 10^{-6})} = \frac{1.5}{13.6 \times 10^3} \approx 1.1 \times 10^{-4}\text{ m/s}$.

Ans: Mobility is defined as the magnitude of drift velocity acquired per unit applied electric field ($\mu = v_d/E$). SI unit is $\text{m}^2\text{V}^{-1}\text{s}^{-1}$. Since $v_d = \frac{eE\tau}{m}$, mobility $\mu = \frac{e\tau}{m}$.

Ans: Resistivity $\rho = \frac{m}{ne^2\tau}$. For semiconductors, as temperature increases, more covalent bonds break, exponentially increasing the charge carrier number density ($n$). Although $\tau$ decreases slightly, the massive increase in $n$ dominates, causing overall resistivity ($\rho$) to decrease rapidly.

Ans: First, find $\alpha$: $R_{100} = R_0(1 + \alpha \times 100) \implies 5.23 = 5.0(1 + 100\alpha) \implies \alpha = \frac{5.23 - 5.0}{5.0 \times 100} = \frac{0.23}{500} = 4.6 \times 10^{-4}\text{ }^\circ\text{C}^{-1}$.
Now find T: $R_T = R_0(1 + \alpha T) \implies 5.795 = 5.0(1 + 4.6 \times 10^{-4} T) \implies \frac{0.795}{5.0} = 4.6 \times 10^{-4} T \implies 0.159 = 4.6 \times 10^{-4} T \implies T = 345.6^\circ\text{C}$.

Ans: Alloys like manganin have: 1. A very low temperature coefficient of resistance ($\alpha \approx 0$), so their resistance barely changes with ambient temperature changes. 2. A relatively high resistivity.

Ans: $R_1 (\text{at } 27^\circ\text{C}) = V/I_1 = 230/3.2 = 71.875\text{ }\Omega$.
$R_2 (\text{at } T) = V/I_2 = 230/2.8 = 82.143\text{ }\Omega$.
Using $R_2 = R_1[1 + \alpha(T - 27)] \implies T - 27 = \frac{R_2 - R_1}{R_1 \alpha} = \frac{82.143 - 71.875}{71.875 \times 1.7 \times 10^{-4}} \approx \frac{10.268}{0.0122} \approx 840^\circ\text{C}$.
$T = 840 + 27 = 867^\circ\text{C}$.

Ans: $R_T = R_0(1 + \alpha T) \implies 2R_0 = R_0(1 + \alpha T) \implies 2 = 1 + \alpha T \implies 1 = \alpha T$.
$T = \frac{1}{\alpha} = \frac{1}{3.9 \times 10^{-3}} = \frac{1000}{3.9} \approx 256.4^\circ\text{C}$.

Ans: $R = \frac{V^2}{P}$. Since $V$ is same ($220\text{V}$), $R \propto 1/P$. The bulb with lower power has higher resistance. Therefore, the $60\text{ W}$ bulb has higher resistance.

Ans: In series, current $I$ is constant. Power dissipated $P = I^2 R$. Since $P \propto R$, the bulb with higher resistance dissipates more power. Therefore, the $60\text{ W}$ bulb glows brighter.

Ans: In parallel, voltage $V$ is constant. Power dissipated $P = V^2 / R$. Since $P \propto 1/R$, the bulb with lower resistance dissipates more power. Therefore, the $100\text{ W}$ bulb glows brighter (as expected in home wiring).

Ans: Power $P = VI = 220 \times 5 = 1100\text{ W} = 1.1\text{ kW}$.
Energy $E = P \times t = 1.1\text{ kW} \times 2\text{ h} = 2.2\text{ kWh}$ (units).

Ans: Time $t = 5 \times 60 = 300\text{ s}$.
Heat $H = \frac{V^2}{R}t = \frac{10^2}{100} \times 300 = \frac{100}{100} \times 300 = 300\text{ Joules}$.

Ans: $P = V^2/R$. If $V' = V/2$, then $P' = (V/2)^2 / R = \frac{V^2/4}{R} = \frac{1}{4} \frac{V^2}{R} = P/4$. Power is reduced to one-fourth.

Ans: EMF is the max potential difference when the cell is in an open circuit (no current). Terminal PD is the potential difference across the cell when it is in a closed circuit (supplying current). EMF is a characteristic of the cell; Terminal PD depends on current drawn.

Ans: $I = \frac{E}{R+r} \implies 0.5 = \frac{10}{R+3} \implies R+3 = 20 \implies R = 17\text{ }\Omega$.

Ans: $V = IR = 0.5 \times 17 = 8.5\text{ V}$. (Alternatively: $V = E - Ir = 10 - 0.5 \times 3 = 10 - 1.5 = 8.5\text{ V}$).

Ans: Terminal voltage is strictly greater than EMF only when the cell is being **charged** by an external external source. Equation: $V = E + Ir$.

Ans: Let terminal voltage be $V$. Current $I = I_1 + I_2$.
$I = (\frac{E_1 - V}{r_1}) + (\frac{E_2 - V}{r_2}) = (\frac{E_1}{r_1} + \frac{E_2}{r_2}) - V(\frac{1}{r_1} + \frac{1}{r_2})$.
Rearranging to form $V = E_{eq} - I r_{eq}$: $V = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} - I \frac{1}{\frac{1}{r_1} + \frac{1}{r_2}}$.
Thus, $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$.

Ans: $E_{eq} = 4 \times 2 = 8\text{ V}$. $r_{eq} = 4 \times 0.2 = 0.8\text{ }\Omega$.
Current $I = \frac{E_{eq}}{R + r_{eq}} = \frac{8}{7.2 + 0.8} = \frac{8}{8.0} = 1\text{ A}$.

Ans: Sum of currents entering a junction equals sum of currents leaving it ($\sum I = 0$). It is based on the Law of Conservation of Charge (charge cannot accumulate at a junction).

Ans: The algebraic sum of changes in potential around any closed loop is zero ($\sum \Delta V = 0$). It is based on the Law of Conservation of Energy (electrostatic force is conservative).

Ans: Tracing clockwise: across $R_1$ is a drop ($-I R_1$); across $E_1$ (positive to negative) is a drop ($-E_1$); across $E_2$ (negative to positive) is a gain ($+E_2$); across $R_2$ is a drop ($-I R_2$).
Equation: $-I R_1 - E_1 + E_2 - I R_2 = 0 \implies E_2 - E_1 = I(R_1 + R_2)$.

Ans: Let node potential be $V$ and other node be $0\text{V}$. Assume all currents leaving.
$I_1 + I_2 + I_3 = 0 \implies \frac{V - 10}{2} + \frac{V - 5}{1} + \frac{V - 0}{5} = 0$.

Ans: Potential can either increase or decrease depending on the direction of traversal relative to current flow and battery polarity. A strict convention ensures that energy gains (rises in potential) perfectly balance energy losses (drops in potential), resulting in a zero sum.

Ans: [Standard Diagram: Diamond with P,Q,R,S and G in middle]. Balanced condition: No current through Galvanometer ($I_g = 0$), which implies $V_B = V_D$. Formula: $\frac{P}{Q} = \frac{R}{S}$.

Ans: Let $I_1$ flow through P and Q, $I_2$ through R and S (since $I_g=0$).
Loop ABDA: $-I_1 P + 0 + I_2 R = 0 \implies I_1 P = I_2 R$.
Loop BCDB: $-I_1 Q - 0 + I_2 S = 0 \implies I_1 Q = I_2 S$.
Divide the two equations: $\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \implies \frac{P}{Q} = \frac{R}{S}$.

Ans: The balance point remains completely unaffected. The bridge remains balanced because interchanging them merely turns the bridge sideways, preserving the proportionality $\frac{P}{R} = \frac{Q}{S}$, which is mathematically identical to $\frac{P}{Q} = \frac{R}{S}$.

Ans: The wire is divided into lengths $l$ and $(100-l)$. Using Wheatstone principle $\frac{R}{S} = \frac{l}{100-l} \implies S = R \left( \frac{100-l}{l} \right)$.

Ans: $\frac{Y}{12.5} = \frac{l}{100-l} = \frac{39.5}{100-39.5} = \frac{39.5}{60.5}$.
$Y = 12.5 \times \frac{39.5}{60.5} \approx 8.16\text{ }\Omega$.

Ans: Thick copper strips have a large cross-sectional area and very low resistivity, ensuring their resistance is negligible. This prevents them from altering the effective resistances in the bridge arms, maintaining accuracy.

Ans: Principle: Potential drop across any length of the wire is directly proportional to that length ($V \propto l$). Conditions: 1. Steady, constant current must flow. 2. The wire must have a uniform cross-sectional area and uniform composition.

Ans: A voltmeter draws a small current from the cell to deflect its pointer, so it measures terminal voltage ($V = E-Ir$). A potentiometer measures at the null point, drawing zero current from the cell, thus measuring true EMF accurately.

Ans: Let potential gradient be $k$. At balance point for cell 1, no current flows, so $E_1$ equals potential drop across $l_1$: $E_1 = k l_1$. Similarly for cell 2: $E_2 = k l_2$. Dividing both equations gives $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.

Ans: $\frac{E_1}{E_2} = \frac{l_1}{l_2} \implies \frac{1.25}{E_2} = \frac{35.0}{63.0}$.
$E_2 = 1.25 \times \frac{63.0}{35.0} = 1.25 \times 1.8 = 2.25\text{ V}$.

Ans: First, balance EMF $E$ on open circuit (length $l_1$, so $E = k l_1$). Then close a shunt resistance $R$ across the cell. Balance the terminal voltage $V$ (length $l_2$, so $V = k l_2$). Internal resistance $r = (\frac{E}{V} - 1)R = \left( \frac{l_1}{l_2} - 1 \right) R$.

Ans: Sensitivity increases by decreasing the potential gradient ($k$). This is achieved by either increasing the length of the potentiometer wire or by decreasing the primary circuit current using a rheostat.

Ans: Connect a very low resistance (Shunt, $S$) in parallel with the Galvanometer coil. This provides a low resistance path for the majority of the current, protecting the sensitive coil and allowing high current measurement.

Ans: Since G and S are in parallel, potential difference is same.
$V_G = V_S \implies I_g \times G = (I - I_g) \times S$.
Therefore, $S = \frac{I_g \times G}{I - I_g}$.

Ans: Connected in series because current must flow *through* it to be measured. Must have very low resistance so it does not significantly increase the total resistance of the circuit, which would alter the very current it is trying to measure.

Ans: Since G and R are in series, the total potential $V$ drops across both.
$V = V_G + V_R = I_g G + I_g R = I_g(G + R)$.
$G + R = \frac{V}{I_g} \implies R = \frac{V}{I_g} - G$.

Ans: Required shunt $S = \frac{I_g G}{I - I_g} = \frac{2.5 \times 10^{-3} \times 30}{7.5 - 0.0025} \approx \frac{0.075}{7.5} = 0.01\text{ }\Omega$. Connect $0.01\text{ }\Omega$ in parallel.

Ans: Required series resistance $R = \frac{V}{I_g} - G = \frac{10}{2.5 \times 10^{-3}} - 30 = 4000 - 30 = 3970\text{ }\Omega$. Connect $3970\text{ }\Omega$ in series.

Ans: If ammeter resistance increases, the total resistance of the circuit increases. By Ohm's law ($I = V/R_{total}$), the current in the circuit will decrease. Thus, the ammeter will read a lower current than the actual value.

Ans: Since the galvanometer $G$ and the high resistance $R$ are connected in series, the effective resistance is simply their sum: $R_v = G + R$.