Vardaan Learning Institute

Ans: Let the equivalent resistance of the infinite ladder be $R_{eq}$. Since it is infinite, removing one repeating unit leaves the resistance unchanged.
Thus, $R_{eq} = R_1 + (R_2 || R_{eq}) = R_1 + \frac{R_2 R_{eq}}{R_2 + R_{eq}}$.
$R_{eq}^2 - R_1 R_{eq} - R_1 R_2 = 0$.
Solving the quadratic equation for $R_{eq}$ gives: $R_{eq} = \frac{R_1 + \sqrt{R_1^2 + 4R_1 R_2}}{2}$ (taking positive root).

Ans: Using symmetric nodal potentials and current distribution:
(a) Face diagonal: Current splits $I \to (I/2, I/4, I/4)$. Equipotential nodes exist. $R_{face} = \frac{3}{4} r$.
(b) Edge: Current splits into parallel paths respecting the reflection symmetry across the diagonal plane. $R_{edge} = \frac{7}{12} r$.

Ans: Note the ratios: $P/Q = 2/4 = 1/2$. $R/S = 4/8 = 1/2$. Wait, the bridge IS balanced! $\frac{AB}{BC} = \frac{2}{4} = \frac{1}{2}$ and $\frac{AD}{DC} = \frac{4}{8} = \frac{1}{2}$.
Since it's balanced, $I_g = 0$. The central $10\text{ }\Omega$ resistor can be removed.
Upper branch $= 2+4=6\text{ }\Omega$. Lower branch $= 4+8=12\text{ }\Omega$.
$R_{eq} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\text{ }\Omega$.

Ans: Use Superposition. Inject current $I$ at node A and extract at infinity. By symmetry, current in each of the 4 branches leaving A is $I/4$.
Now inject $I$ at infinity and extract at adjacent node B. Current in the 4 branches entering B is $I/4$.
Superpose: Inject at A, extract at B. Current in branch AB is $I/4 + I/4 = I/2$.
Voltage drop $V_{AB} = (I/2)R$. $R_{eq} = V_{AB}/I = R/2$.

Ans: In steady state, a DC capacitor acts as an open circuit for its *own* branch, but current flows through the leakage resistance $R_L$.
Total steady current $I = \frac{V}{R + R_L}$.
Voltage across capacitor = Voltage across $R_L = I \times R_L = V \frac{R_L}{R + R_L}$.
Charge $Q = C V_C = \frac{C V R_L}{R + R_L}$.

Ans: Swapping them means the battery is across BD and galvanometer across AC.
The new arms ratio for balance requires $\frac{P}{R} = \frac{Q}{S}$. This is mathematically identical to the original condition $\frac{P}{Q} = \frac{R}{S}$. The bridge remains balanced.

Ans: Let central node potential be $V$. By KCL: $\sum I = 0 \implies \frac{V-10}{1} + \frac{V-20}{2} + \frac{V-30}{3} = 0$.
Multiply by 6: $6(V-10) + 3(V-20) + 2(V-30) = 0$.
$6V - 60 + 3V - 60 + 2V - 60 = 0 \implies 11V = 180 \implies V = \frac{180}{11}\text{ V} \approx 16.36\text{ V}$.

Ans: The two semicircular halves of the ring are in parallel with the diametrical wire.
Resistance of each semicircle $= R_0/2$.
$1/R_{eq} = \frac{1}{R_0/2} + \frac{1}{R_0/2} + \frac{1}{R_d} = \frac{4}{R_0} + \frac{1}{R_d}$.
$R_{eq} = \frac{R_0 R_d}{4R_d + R_0}$.

Ans: Terminal voltage $V = E - Ir$. This is of the form $y = mx + c$ (a straight line with slope $-r$ and intercept $E$).
Power in $r$ is $P_r = I^2 r$. It is maximum when $I$ is maximum, which occurs when external load $R = 0$ (short circuit). $I_{max} = E/r$. Max power in $r = (E/r)^2 r = E^2/r$.

Ans: Let vertices be A, B, C, D. Connect battery across A and B.
By symmetry, nodes C and D are equipotential ($V_C = V_D$).
The resistor between C and D carries no current and can be removed.
We are left with resistor AB ($R$) in parallel with branch ACB ($2R$) and ADB ($2R$).
$1/R_{eq} = 1/R + 1/(2R) + 1/(2R) = 1/R + 1/R = 2/R$. Thus, $R_{eq} = R/2$.

Ans: Consider a disc element of thickness $dx$ at distance $x$. Radius $r(x) = a + (\frac{b-a}{L})x$.
$dR = \frac{\rho dx}{\pi r^2} = \frac{\rho dx}{\pi [a + (\frac{b-a}{L})x]^2}$.
Integrate from $x=0$ to $L$: let $u = a + (\frac{b-a}{L})x$, $du = (\frac{b-a}{L})dx$.
$R = \int_a^b \frac{\rho}{\pi u^2} \frac{L}{b-a} du = \frac{\rho L}{\pi(b-a)} \left[ -\frac{1}{u} \right]_a^b = \frac{\rho L}{\pi(b-a)} (\frac{1}{a} - \frac{1}{b}) = \frac{\rho L}{\pi a b}$.

Ans: Current $I$ is constant. $I = n e A(x) v_d(x)$.
Area $A(x) = \pi [r(x)]^2 = \pi [a + (\frac{b-a}{L})x]^2$.
$v_d(x) = \frac{I}{n e \pi [a + (\frac{b-a}{L})x]^2}$.

Ans: Consider a spherical shell element of radius $r$ and thickness $dr$. Area $A = 4\pi r^2$.
$dR = \rho \frac{dr}{4\pi r^2}$.
$R = \int_{r_1}^{r_2} \frac{\rho dr}{4\pi r^2} = \frac{\rho}{4\pi} \left[ -\frac{1}{r} \right]_{r_1}^{r_2} = \frac{\rho}{4\pi} \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.

Ans: Consider a cylindrical shell element of radius $r$, thickness $dr$, length $L$. Area $A = 2\pi r L$.
$dR = \rho \frac{dr}{2\pi r L}$.
$R = \int_a^b \frac{\rho dr}{2\pi L r} = \frac{\rho}{2\pi L} [\ln r]_a^b = \frac{\rho}{2\pi L} \ln\left(\frac{b}{a}\right)$.

Ans: $I = \int \vec{J} \cdot d\vec{A}$. For a cylindrical wire, $dA = 2\pi r dr$.
$I = \int_0^R J_0(1 - r/R) (2\pi r) dr = 2\pi J_0 \int_0^R (r - r^2/R) dr = 2\pi J_0 [\frac{r^2}{2} - \frac{r^3}{3R}]_0^R$.
$I = 2\pi J_0 (\frac{R^2}{2} - \frac{R^2}{3}) = 2\pi J_0 (\frac{R^2}{6}) = \frac{\pi J_0 R^2}{3}$.

Ans: As $T$ increases due to Joule heating, the term $\beta/T$ decreases, so $e^{\beta/T}$ decreases. Thus, resistivity $\rho$ decreases, and Resistance $R$ decreases.
Since $V$ is constant, Current $I = V/R$ will **increase**.
This causes more heating (Positive feedback/Thermal runaway), leading to potential structural failure (melting) unless limited by external circuitry.

Ans: $J = \sigma E$. From the given equation, $J = e(n\mu_e + p\mu_h)E \implies \sigma = e(n\mu_e + p\mu_h)$.
For an intrinsic semiconductor, $n = p = n_i$ (intrinsic carrier concentration).
Thus, $\sigma_i = e n_i (\mu_e + \mu_h)$.

Ans: Equation of motion: $m \frac{dv}{dt} = -eE_0 \cos(\omega t)$.
$\int dv = \int -\frac{eE_0}{m} \cos(\omega t) dt \implies v(t) = -\frac{eE_0}{m\omega} \sin(\omega t)$.
Amplitude of velocity $v_0 = \frac{eE_0}{m\omega}$.

Ans: Thermal energy: $\frac{1}{2}mv_{th}^2 = \frac{3}{2}kT \implies v_{th} = \sqrt{\frac{3kT}{m}}$.
$v_{th} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{9.1 \times 10^{-31}}} \approx 1.17 \times 10^5\text{ m/s}$.
Typical drift speed is $\sim 10^{-3}\text{ m/s}$. The random thermal speed is roughly $10^8$ times greater than the directed drift speed.

Ans: Consider element $dx$. $dR = \rho(x) \frac{dx}{A} = \frac{\rho_0(1 + \alpha x)}{A} dx$.
$R = \int_0^L \frac{\rho_0}{A}(1 + \alpha x) dx = \frac{\rho_0}{A} [x + \frac{\alpha x^2}{2}]_0^L = \frac{\rho_0}{A} \left( L + \frac{\alpha L^2}{2} \right)$.

Ans: Effective resistance of the wire portion AB is $R_{eff} = \frac{R_w R_s}{R_w + R_s}$.
Total primary circuit resistance $R_{total} = r_0 + R_h + R_{eff}$.
Main current $I = \frac{E_0}{r_0 + R_h + R_{eff}}$.
Potential drop across AB is $V_{AB} = I \times R_{eff} = \frac{E_0 \cdot R_{eff}}{r_0 + R_h + R_{eff}}$.
Potential gradient $k = \frac{V_{AB}}{L} = \frac{E_0}{L} \cdot \frac{R_w R_s / (R_w+R_s)}{r_0 + R_h + \frac{R_w R_s}{R_w+R_s}}$.

Ans: Resistance of left part $R_L = \int_0^l \lambda_0(1+\alpha x) dx = \lambda_0(l + \alpha l^2/2)$.
Resistance of right part $R_R = \int_l^{100} \lambda_0(1+\alpha x) dx = \lambda_0[(100-l) + \frac{\alpha}{2}(100^2 - l^2)]$.
Balanced condition: $P/Q = R_L / R_R = \frac{l + \alpha l^2/2}{100-l + \frac{\alpha}{2}(10000 - l^2)}$.

Ans: True voltage $V_{true} = E \frac{R_1}{R_1+R_2}$.
With voltmeter $R_v$, equivalent resistance $R_{eq} = \frac{R_1 R_v}{R_1+R_v}$.
Measured voltage $V_{meas} = E \frac{R_{eq}}{R_{eq}+R_2}$. Since $R_{eq} < R_1$, $V_{meas} < V_{true}$.
Fractional error $= \frac{V_{true} - V_{meas}}{V_{true}}$, which evaluates to $\approx \frac{R_1 R_2}{R_v(R_1+R_2)}$ for large $R_v$.

Ans: Let $X, Y$ be resistances. Let $\sigma$ be resistance/cm. $l_1$ is balance point.
$X/Y = (l_1\sigma + e_1) / ((100-l_1)\sigma + e_2)$.
Interchanging $X, Y$ gives new balance $l_2$: $Y/X = (l_2\sigma + e_1) / ((100-l_2)\sigma + e_2)$.
Using properties of proportions, it yields $X - Y = \sigma(l_2 - l_1)$.

Ans: Required $k = 1\text{ }\mu\text{V/mm} = 10^{-3}\text{ V/m}$.
$V_{AB} = k \times L = 10^{-3} \times 10 = 0.01\text{ V}$.
Current in wire $I = V_{AB}/R_w = 0.01 / 20 = 5 \times 10^{-4}\text{ A}$.
Also $I = \frac{E}{R + R_w} \implies 5 \times 10^{-4} = \frac{5}{R + 20}$.
$R + 20 = 10000 \implies R = 9980\text{ }\Omega$.

Ans: Without $R$, it balances $E_1 = k l_1$.
With $R$, it balances terminal voltage $V = k l_2$.
Since $V = E_1 \frac{R}{R+r_1}$, we get $k l_2 = k l_1 \frac{R}{R+r_1} \implies \frac{l_1}{l_2} = \frac{R+r_1}{R} = 1 + \frac{r_1}{R}$.
$r_1 = R \left( \frac{l_1}{l_2} - 1 \right)$.

Ans: Pass the unknown current $I$ through a known standard resistance $R_{std}$ in the secondary circuit. Use the potentiometer to measure the potential drop $V_{std}$ across this standard resistor. By balancing, $V_{std} = k l$. Then calculate current as $I = V_{std} / R_{std} = (k l) / R_{std}$.

Ans: Total resistance $R_{tot} = r + R_A + R_V$.
The ammeter reads the current: $I = \frac{E}{r + R_A + R_V}$.
The voltmeter reads the potential drop across itself: $V = I R_V = E \frac{R_V}{r + R_A + R_V}$. (Since $R_V$ is typically very large, it will read close to $E$).

Ans: "End error" accounts for the resistance of the thick copper strips and soldered joints at ends A and B, expressed as equivalent lengths of the bridge wire ($e_1, e_2$).
Modified condition: $\frac{P}{Q} = \frac{l + e_1}{(100-l) + e_2}$.

Ans: Deflection $\theta = \frac{NAB}{k}I$.
Current Sensitivity $I_s = \theta/I = \frac{NAB}{k}$. Increasing $N$ increases $I_s$ proportionally.
Voltage Sensitivity $V_s = \theta/V = \frac{\theta}{IR} = \frac{NAB}{kR}$. When $N$ is doubled, length of wire doubles, so Resistance $R$ doubles. Thus, $V_s$ remains **unchanged**.

Ans: Equivalent EMF $E_{eq} = nE$. Equivalent internal resistance $r_{eq} = nr/m$.
Current $I = \frac{nE}{R + nr/m}$. Power $P = I^2 R = \frac{n^2 E^2 R}{(R + nr/m)^2}$.
By max power transfer theorem, $P$ is max when external resistance equals equivalent internal resistance. $\implies R = nr/m$.

Ans: $H = \int_0^{t_0} I^2 R dt = R \int_0^{t_0} I_0^2 (1 - t/t_0)^2 dt$.
Let $u = 1 - t/t_0$, $du = -1/t_0 dt$.
$H = -I_0^2 R t_0 \int_1^0 u^2 du = I_0^2 R t_0 [\frac{u^3}{3}]_0^1 = \frac{I_0^2 R t_0}{3}$.

Ans: Heat generated $P_{in} = I^2 R = I^2 (\rho \frac{L}{\pi r^2})$.
Heat radiated $P_{out} = (\text{const}) \times \text{Surface Area} = h (2\pi r L)$.
At steady melting temp: $I^2 \rho \frac{L}{\pi r^2} = h 2\pi r L$. Notice $L$ cancels out.
$I^2 = \frac{2\pi^2 h}{\rho} r^3 \implies I \propto r^{3/2}$.

Ans: For series combination of appliances rated for the same voltage $V$, the equivalent power is given by $\frac{1}{P_{eq}} = \frac{1}{P_1} + \frac{1}{P_2}$.
$P_{eq} = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} = 66.67\text{ W}$.

Ans: Total power input $P_{in} = VI$. Power lost as heat $P_{heat} = I^2 r$.
Mechanical power output $P_{mech} = P_{in} - P_{heat} = VI - I^2 r$.
Efficiency $\eta = \frac{P_{mech}}{P_{in}} = \frac{VI - I^2 r}{VI} = 1 - \frac{Ir}{V}$.

Ans: Current $I(t) = \frac{V_0}{R} e^{-t/RC}$.
$H = \int_0^\infty I^2 R dt = R \int_0^\infty (\frac{V_0}{R})^2 e^{-2t/RC} dt = \frac{V_0^2}{R} \left[ \frac{e^{-2t/RC}}{-2/RC} \right]_0^\infty$.
$H = \frac{V_0^2}{R} (0 - \frac{1}{-2/RC}) = \frac{V_0^2}{R} \frac{RC}{2} = \frac{1}{2} C V_0^2$.

Ans: $E_{eq} = \frac{10/1 + 8/2}{1/1 + 1/2} = \frac{10+4}{1.5} = \frac{14}{1.5} = 28/3\text{ V}$.
$r_{eq} = \frac{1 \times 2}{1+2} = 2/3\text{ }\Omega$.
Current $I = \frac{E_{eq}}{R + r_{eq}} = \frac{28/3}{5 + 2/3} = \frac{28/3}{17/3} = 28/17\text{ A}$.
Power $P = I^2 R = (28/17)^2 \times 5 = (784 / 289) \times 5 \approx 13.56\text{ W}$.

Ans: Original $P = V^2/R$. Cutting in half makes new resistance $R' = R/2$.
New power $P' = V^2 / (R/2) = 2(V^2/R) = 2P$.
Power is doubled. Percentage change $= \frac{2P - P}{P} \times 100\% = +100\%$.

Ans: Two resistors of $R/2$ in parallel: $R_{eq} = \frac{(R/2)(R/2)}{R/2 + R/2} = R/4$.
New power $P'' = V^2 / (R/4) = 4 (V^2/R) = 4P$. Power is four times the original.

Ans: [Graph: P on Y-axis, R on X-axis. Curve starts at origin, rises to a single smooth peak, and then decays asymptotically towards zero].
The peak coordinates are exactly at $(R=r, P_{max}=E^2/4r)$.

Ans: The circle is divided in ratio $60^\circ : 300^\circ$, which is $1:5$. Resistance divides as $1:5$. Total $36 = R_1 + R_2 \implies R_1 = 6\text{ }\Omega, R_2 = 30\text{ }\Omega$. They are in parallel. $R_{eq} = \frac{6 \times 30}{6 + 30} = \frac{180}{36} = 5$. 5

Ans: Check ratio: $P/Q = 10/20 = 1/2$. $R/S = 15/30 = 1/2$. Bridge is perfectly balanced. Current through galvanometer is exactly zero. 0

Ans: Total resistance $= 20 + 5 + 15 = 40\text{ }\Omega$. Current $I = 6 / 40 = 0.15\text{ A}$. $V_{wire} = I R_{wire} = 0.15 \times 20 = 3.0\text{ V}$. Gradient $k = V/L = 3.0\text{V} / 10\text{m} = 0.3\text{ V/m} = 0.3 \times 1000\text{ mV} / 100\text{ cm} = 3\text{ mV/cm}$. 3

Ans: $I = \int_0^R J(2\pi r dr) = 2\pi(10^5) \int_0^{2\times 10^{-3}} r^3 dr = 2\pi(10^5) [\frac{r^4}{4}]_0^{0.002} = \frac{\pi}{2} 10^5 (16 \times 10^{-12}) = 8\pi \times 10^{-7} = 0.8\pi \times 10^{-6} \approx 2.512 \times 10^{-6}\text{ A}$. $X = 2.512$. 2.512

Ans: $S = \frac{I_g G}{I_0 - I_g} \implies 0.1 = \frac{0.01 \times 100}{I_0 - 0.01} \implies 0.1(I_0 - 0.01) = 1 \implies I_0 - 0.01 = 10 \implies I_0 = 10.01\text{ A}$. 10.01

Ans: $R_{eq}$ across body diagonal $= \frac{5}{6} r = \frac{5}{6} (1) = 5/6\text{ }\Omega$. Current $I = V / R_{eq} = 2.5 / (5/6) = 2.5 \times 6 / 5 = 3\text{ A}$. 3

Ans: Charging current $I = \frac{110 - 50}{28 + 2} = \frac{60}{30} = 2\text{ A}$. Terminal voltage during charging $V = E + Ir = 50 + (2)(2) = 54\text{ V}$. 54

Ans: Let resistors be $x, y$. $S = x+y$, $P = xy/(x+y)$. $S/P = (x+y)^2 / xy$. Let $y/x = k$. Ratio is $(k+1)^2/k = k + 2 + 1/k$. Min value (by AM-GM on $k+1/k \ge 2$) is $2+2=4$ (when $x=y$). 4

Ans: $E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{1.5/0.2 + 2.0/0.3}{1/0.2 + 1/0.3} = \frac{7.5 + 6.66}{5 + 3.33} = \frac{7.5 + 20/3}{5 + 10/3} = \frac{42.5/3}{25/3} = \frac{42.5}{25} = 1.7\text{ V}$. 1.7

Ans: Value $= 10 \times 10^2 = 1000\text{ }\Omega$. Tolerance $\pm 10\%$. Max $R = 1000 + 10\%(1000) = 1000 + 100 = 1100\text{ }\Omega$. 1100

Ans: $\frac{R_{left}}{R_{right}} = \frac{l}{100-l} \implies \frac{15}{S} = \frac{40}{60} = \frac{2}{3} \implies S = 15 \times 3 / 2 = 22.5\text{ }\Omega$. 22.5

Ans: Series power $P_{eq} = \frac{P_1 P_2}{P_1 + P_2} = \frac{1000 \times 2000}{3000} = \frac{2000}{3} \approx 666.67\text{ W}$. 666.67

Ans: $R \propto l^2$. For small percentage changes, $\%\Delta R = 2 \times (\%\Delta l) = 2 \times 0.2\% = 0.4\%$. 0.4

Ans: $I = \frac{E}{10r+r} = \frac{E}{11r}$. $V_L = I(10r) = \frac{10}{11}E$. Gradient $k = V_L/L = \frac{10E}{11L}$. Balance: $E/2 = kl = (\frac{10E}{11L})l \implies l/L = \frac{11}{20} = 0.55$. 0.55

Ans: $J = I/A = \sigma E \implies E = \rho I / A$. If $A' = A/2$, $E' = \rho I / (A/2) = 2(\rho I / A) = 2E$. So $n = 2$. 2

Ans: $H = \frac{V^2}{R}t \implies t = \frac{HR}{V^2}$. $t \propto 1/V^2$. $t' = t (V/V')^2 = 10 (1 / 0.8)^2 = 10 (1.25)^2 = 10 (1.5625) = 15.625\text{ mins}$. 15.625

Ans: Standard result (derived via superposition in Section A, Q4) is $R_{eq} = R/2 = 1/2 = 0.5\text{ }\Omega$. 0.5

Ans: $R_T = R_0(1 + \alpha \Delta T) \implies 2 = 1(1 + 0.00125 \Delta T) \implies 1 = 0.00125 \Delta T \implies \Delta T = 1/0.00125 = 800$. $T = 300 + 800 = 1100\text{ K}$. 1100

Ans: $P = V^2/R \implies P \propto V^2$. $V' = 1.1 V$. $P' \propto (1.1 V)^2 = 1.21 V^2 = 1.21 P$. Increase is $0.21 P$, which is $21\%$. 21

Ans: $V = E - Ir \implies 10 = 12 - 50r \implies 50r = 2 \implies r = 2/50 = 0.04\text{ }\Omega$. 0.04