7.Wire stretched to increase length by $10\%$. $\%$ change in $R$.
Ans: When stretched, volume $V$ is constant. $R = \rho l/A = \rho l^2/V \implies R \propto l^2$.
New length $l' = l + 0.1l = 1.1l$.
New resistance $R' \propto (1.1l)^2 = 1.21 l^2 = 1.21R$.
Change in $R = 1.21R - R = 0.21R$. Percentage change $= 21\%$.
8.V-I graph for $T_1$ (steeper) and $T_2$. Which is higher?
Ans: In a V-I graph (V on Y-axis, I on X-axis), slope $= V/I = R$. Since $T_1$ is steeper, slope of $T_1 >$ slope of $T_2$, implying $R_1 > R_2$. For metallic conductors, resistance increases with temperature. Therefore, $T_1 > T_2$.
9.Carbon block $a \times b \times c$ ($a
Ans: $R = \rho l/A$.
For maximum $R$, length $l$ must be max ($c$) and Area $A$ must be min ($a \times b$). So, apply across faces separated by distance $c$.
For minimum $R$, length $l$ must be min ($a$) and Area max ($b \times c$). Apply across faces separated by distance $a$.
10.Resistance of hollow cylinder (longitudinal current).
Ans: Current flows along length $L$. The cross-sectional area available for conduction is the area of the annular ring: $A = \pi r_2^2 - \pi r_1^2 = \pi(r_2^2 - r_1^2)$.
Resistance $R = \rho \frac{L}{A} = \frac{\rho L}{\pi(r_2^2 - r_1^2)}$.
11.Define conductance and conductivity. Write SI units.
Ans: Conductance ($G$) is the reciprocal of resistance ($G = 1/R$). SI unit: Siemens ($\text{S}$) or $\Omega^{-1}$. Conductivity ($\sigma$) is the reciprocal of resistivity ($\sigma = 1/\rho$). SI unit: $\text{S m}^{-1}$ or $\Omega^{-1}\text{m}^{-1}$.
12.V-I characteristic for non-ohmic device (GaAs). Mark regions.
Ans: [Student should draw a curve starting linearly from origin, reaching a peak, and then dipping downwards before rising again].
The initial straight portion is the **Linear region**. The downward-sloping region where current decreases as voltage increases is the **Negative resistance region**.
13.Define mobility. Prove $\sigma = ne\mu_e$.
Ans: Mobility ($\mu$) is the drift velocity per unit electric field ($\mu = v_d/E$).
We know $I = neAv_d \implies J = I/A = nev_d$.
From Ohm's law, $J = \sigma E$. Therefore, $\sigma E = ne v_d$.
$\sigma = ne (v_d/E)$. Substituting $\mu_e = v_d/E$ gives $\sigma = ne\mu_e$.
14.Two wires series. $L_1/L_2 = 1/2$, $r_1/r_2 = 1/2$. Compare $v_d$.
Ans: In series, current $I$ is constant. $v_d = \frac{I}{neA} \implies v_d \propto \frac{1}{A} \propto \frac{1}{r^2}$.
$\frac{v_{d1}}{v_{d2}} = \left(\frac{r_2}{r_1}\right)^2 = \left(\frac{2}{1}\right)^2 = 4:1$. (Length does not affect drift velocity in series circuit directly, only radius does).
15.Temperature increases. How do $\tau, v_d, n$ change for a metal?
Ans: (a) Relaxation time ($\tau$) decreases due to increased frequent collisions with vibrating lattice ions.
(b) Drift velocity ($v_d \propto \tau$) also decreases (assuming $E$ is constant).
(c) Electron density ($n$) remains practically unchanged for a metal.
16.Time to drift $3.0\text{ m}$. $n=8.5 \times 10^{28}, A=2.0 \times 10^{-6}, I=3.0\text{ A}$.
Ans: $v_d = \frac{I}{neA} = \frac{3.0}{(8.5 \times 10^{28})(1.6 \times 10^{-19})(2 \times 10^{-6})} \approx 1.1 \times 10^{-4}\text{ m/s}$.
Time $t = \frac{\text{Distance}}{v_d} = \frac{3.0}{1.1 \times 10^{-4}} \approx 2.7 \times 10^4\text{ s}$ (about 7.5 hours).
17.Why does a bulb light up instantly despite low $v_d$?
Ans: When the switch is closed, an electric field is established throughout the entire circuit almost instantaneously (at the speed of light). This field exerts a force on all free electrons everywhere in the circuit simultaneously, causing them all to drift together instantly, thus establishing the current immediately.
18.Relation $\vec{J}$ and $\vec{v}_d$. Parallel or antiparallel?
Ans: $I = neAv_d \implies J = nev_d$. In vector form, taking charge sign into account: $\vec{J} = -ne\vec{v}_d$ for electrons. Since electrons carry negative charge, their drift velocity $\vec{v}_d$ is exactly **antiparallel** to the current density vector $\vec{J}$ (which follows conventional positive current direction).
19.Why is Nichrome preferred for standard wire-bound resistors?
Ans: Based on the graph, Nichrome has a very small temperature coefficient of resistivity (nearly flat slope). Its resistance changes negligibly with variations in temperature (Joule heating or ambient), ensuring the standard resistance value remains highly stable and accurate.
20.Condition for temperature independent equivalent resistance ($R_1, \alpha_1$ and $R_2, \alpha_2$ in series).
Ans: $R_{eq} = R_1(1+\alpha_1\Delta T) + R_2(1+\alpha_2\Delta T) = (R_1+R_2) + (R_1\alpha_1 + R_2\alpha_2)\Delta T$.
For $R_{eq}$ to be independent of $\Delta T$, the coefficient of $\Delta T$ must be zero.
$R_1\alpha_1 + R_2\alpha_2 = 0$. This implies one $\alpha$ must be negative (like a semiconductor or carbon) to compensate for the metal.
21.At what temperature is resistance twice $R_0$?
Ans: $R_T = R_0(1 + \alpha T)$. We want $R_T = 2R_0$.
$2R_0 = R_0(1 + \alpha T) \implies 2 = 1 + \alpha T \implies 1 = \alpha T \implies T = 1/\alpha$.
22.Microscopic reason for semiconductor $\rho$ decrease with T.
Ans: $\rho = \frac{m}{ne^2\tau}$. As T increases, $\tau$ decreases slightly due to collisions. However, thermal energy breaks many covalent bonds, releasing exponentially more charge carriers, causing $n$ to increase drastically. The massive increase in $n$ overpowers the small decrease in $\tau$, causing overall resistivity $\rho$ to drop.
23.$R_{27} = 100\text{ }\Omega$, $R_T = 117\text{ }\Omega$, $\alpha = 1.7 \times 10^{-4}$. Find T.
Ans: $R_T = R_{27} [1 + \alpha(T - 27)] \implies 117 = 100 [1 + 1.7 \times 10^{-4}(T - 27)]$.
$1.17 = 1 + 1.7 \times 10^{-4}(T - 27) \implies 0.17 = 1.7 \times 10^{-4}(T - 27)$.
$T - 27 = \frac{0.17}{1.7 \times 10^{-4}} = \frac{10^4}{10} = 1000$.
$T = 1000 + 27 = 1027^\circ\text{C}$.
24.Practical application of highly negative $\alpha$ thermistor.
Ans: Thermistors are widely used as highly sensitive electronic **temperature sensors** or digital thermometers because a tiny change in temperature causes a large, easily measurable drop in resistance. They are also used for inrush current limiting in power supplies.
25.Maximum Power Transfer Theorem derivation.
Ans: Power $P = I^2 R = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2}$.
For max power, $dP/dR = 0$. Using quotient rule: $\frac{(R+r)^2 E^2 - E^2 R [2(R+r)]}{(R+r)^4} = 0$.
$(R+r)^2 - 2R(R+r) = 0 \implies (R+r)(R+r - 2R) = 0 \implies r - R = 0 \implies R = r$.
Maximum power is transferred when external load $R$ equals internal resistance $r$.
26.Equivalent power of $P_1, P_2$ in series across same V.
Ans: Resistances: $R_1 = V^2/P_1$ and $R_2 = V^2/P_2$.
In series, $R_{eq} = R_1 + R_2 = \frac{V^2}{P_1} + \frac{V^2}{P_2}$.
Equivalent power $P_{eq} = \frac{V^2}{R_{eq}} = \frac{V^2}{\frac{V^2}{P_1} + \frac{V^2}{P_2}} = \frac{1}{\frac{1}{P_1} + \frac{1}{P_2}}$.
Thus, $P_{eq} = \frac{P_1 P_2}{P_1 + P_2}$.
27.Coils boiling water: $t_1, t_2$. Find time in parallel.
Ans: Heat $H$ is constant. Power $P = H/t$.
In parallel, $P_{eq} = P_1 + P_2 \implies \frac{H}{t_p} = \frac{H}{t_1} + \frac{H}{t_2} \implies \frac{1}{t_p} = \frac{1}{t_1} + \frac{1}{t_2}$.
Time in parallel $t_p = \frac{t_1 t_2}{t_1 + t_2}$.
28.Time in series for Q27 coils.
Ans: In series, $P_{eq} = \frac{P_1 P_2}{P_1 + P_2} \implies \frac{H}{t_s} = \frac{H/t_1 \cdot H/t_2}{H/t_1 + H/t_2} = \frac{H^2 / (t_1 t_2)}{H(t_1+t_2)/(t_1 t_2)} = \frac{H}{t_1+t_2}$.
Thus, $\frac{1}{t_s} = \frac{1}{t_1+t_2} \implies t_s = t_1 + t_2$.
29.Bulb $100\text{W}, 220\text{V}$ on $110\text{V}$. Actual power?
Ans: Resistance $R = \frac{V_{rated}^2}{P_{rated}} = \frac{220^2}{100} = 484\text{ }\Omega$.
Actual Power $P_{actual} = \frac{V_{actual}^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\text{ W}$.
(Shortcut: Voltage halved $\implies$ Power becomes $(1/2)^2 = 1/4$th. $100/4 = 25\text{ W}$).
30.Efficiency of cell delivering maximum power.
Ans: Max power condition: $R = r$.
Power delivered to load (useful) $P_{out} = I^2 R$.
Total power generated by cell $P_{total} = I^2(R+r) = I^2(2R)$.
Efficiency $\eta = \frac{P_{out}}{P_{total}} \times 100\% = \frac{I^2 R}{2 I^2 R} \times 100\% = 50\%$.
31.What do Y-intercept, X-intercept, and slope represent in V-I graph?
Ans: Equation is $V = E - Ir$, which is $y = mx + c \implies V = (-r)I + E$.
(a) Y-intercept ($I=0$): represents the true EMF ($E$) of the cell.
(b) X-intercept ($V=0$): represents the Short Circuit Current ($I_{max} = E/r$).
(c) Negative slope: The magnitude of the slope represents the Internal Resistance ($r$) of the cell.
32.Cell short-circuited. Find V and I.
Ans: Short-circuit means external resistance $R = 0$.
Current $I = \frac{E}{R+r} = \frac{E}{0+r} = \frac{E}{r}$ (maximum current).
Terminal voltage $V = IR = I \times 0 = 0\text{ V}$.
33.Equivalent EMF of parallel cells. Condition for $E_{eq} = E_1$.
Ans: Equivalent EMF $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$.
Condition: $\frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = E_1 \implies E_1 r_2 + E_2 r_1 = E_1 r_1 + E_1 r_2 \implies E_2 r_1 = E_1 r_1 \implies E_2 = E_1$.
34.$N$ cells in series, $n$ reversed. Total EMF?
Ans: Total number of properly connected cells is $(N - n)$.
The $n$ reversed cells actively oppose the $(N - n)$ cells.
$E_{eq} = (N - n)E - nE = (N - 2n)E$. (Note: Internal resistance remains $Nr$, it doesn't subtract).
35.Condition for max current in mixed grouping ($m$ rows, $n$ cells).
Ans: Total cells = $mn$. EMF of one row = $nE$. Total internal resistance of one row = $nr$.
Since $m$ rows are in parallel, $r_{eq} = \frac{nr}{m}$. $E_{eq} = nE$ (parallel voltage remains same).
Current $I = \frac{nE}{R + nr/m} = \frac{mnE}{mR + nr}$.
For $I$ to be maximum, denominator must be minimum. By AM-GM, $mR + nr$ is min when $mR = nr \implies R = \frac{nr}{m}$. (External resistance = Equivalent internal resistance).
36.Why is V > E during charging?
Ans: During charging, current $I$ is forced *into* the positive terminal by an external charger. The external source must overcome both the cell's opposing EMF ($E$) and the potential drop across its internal resistance ($Ir$). Therefore, applied voltage $V = E + Ir$, making $V > E$.
37.Apply Kirchhoff's rules to find current in central resistor.
Ans: Let $I_1$ leave $E_1$ (top branch), $I_2$ leave $E_2$ (bottom branch). They meet at junction A. Current down middle branch $R_3$ is $I_3 = I_1 + I_2$ (Junction Rule).
Left Loop (clockwise): $10 - 2I_1 - 3(I_1 + I_2) = 0 \implies 5I_1 + 3I_2 = 10$.
Right Loop (counter-clockwise): $5 - 1I_2 - 3(I_1 + I_2) = 0 \implies 3I_1 + 4I_2 = 5$.
Solving gives $I_1 = 25/11\text{ A}$, $I_2 = -5/11\text{ A}$. $I_3 = I_1 + I_2 = 20/11\text{ A}$.
38.Role of steady-state in Junction rule.
Ans: Junction rule ($\sum I = 0$) relies on charge conservation. If the current is not steady (e.g., a capacitor is charging at the node), charge would accumulate or deplete over time, invalidating $\sum I_{in} = \sum I_{out}$. Steady-state ensures zero charge accumulation.
39.Branch with $R=5\text{ }\Omega$, $E=12\text{V}$ opposing. $I=2\text{A}$ from A to B. Find $V_A - V_B$.
Ans: Traverse from A to B: $V_A - IR - E = V_B$ (since battery opposes, traverse from + to -, a drop).
$V_A - (2)(5) - 12 = V_B \implies V_A - 10 - 12 = V_B \implies V_A - V_B = 22\text{ V}$.
40.Can Loop rule be applied to open circuit?
Ans: Yes. An open circuit can be mathematically completed by defining the potential difference $V_{AB}$ across the open terminals. The loop equation becomes: Sum of potential drops across components + $V_{AB} = 0$. This is exactly how we calculate open-terminal voltage.
41.Unbalanced vs balanced Wheatstone bridge current distribution.
Ans: In a balanced bridge, the potential at the two central nodes is identical ($V_B = V_D$), so by Ohm's law, zero current flows through the galvanometer. If unbalanced, $V_B \neq V_D$, creating a potential difference that drives a cross-current $I_g$. The **Junction Rule** dictates that this $I_g$ alters the currents in the adjoining arms (e.g., $I_1$ splits into $I_3$ and $I_g$).
42.Skeleton cube resistance across longest body diagonal.
Ans: By symmetry, a current $I$ entering a corner splits into three equal paths ($I/3$). At the next junctions, they split again into ($I/6$). Recombining at the exit corner, they form ($I/3$) again.
$V_{drop} = I_{path1}R + I_{path2}R + I_{path3}R = (I/3)R + (I/6)R + (I/3)R = \frac{5}{6} IR$.
Since $V = I R_{eq}$, $R_{eq} = \frac{5}{6} R$.
43.S shunted by S. Balance point shift?
Ans: Shunting $S$ with $S$ (in parallel) makes the new resistance $S' = S/2$.
Original state: $\frac{R}{S} = \frac{l}{100-l}$. New state: The right gap resistance has halved. To maintain ratio, the ratio $\frac{l}{100-l}$ must also increase (since $\frac{R}{S/2} = 2 \frac{R}{S}$). Thus, $l$ must increase. The balance point shifts to the **right** (towards 100 mark).
44.Derive balanced condition using Kirchhoff rules.
Ans: (See Topic 3.8 Level 1 Solution Q42 for detailed derivation steps. $I_g=0$, apply Loop rule to ABDA and BCDB, equate potential drops $I_1 P = I_2 R$ and $I_1 Q = I_2 S$, and divide).
45.Why accuracy poor at extreme ends?
Ans: At extreme ends (e.g., $l=5\text{ cm}$), the ratio $l / (100-l)$ becomes very skewed. End errors (resistance of thick copper strips and contact resistances) become comparable to the small resistance of the $5\text{ cm}$ wire segment, causing massive percentage errors in the calculation.
46.What are "end errors" and how to minimize?
Ans: End errors arise from the unmeasured resistance of the copper binding strips and the soldering contacts at ends A and B. They are minimized by interchanging $R$ and $S$, finding the new balance point, and taking the mean of the two calculated values of the unknown resistance.
47.$l_1 = 33.7\text{ cm}$. $S$ shunted with $12\text{ }\Omega \implies l_2 = 51.9\text{ cm}$. Find $R, S$.
Ans: Case 1: $\frac{R}{S} = \frac{33.7}{66.3} \approx 0.508$.
Case 2: New $S' = \frac{12S}{12+S}$. $\frac{R}{S'} = \frac{51.9}{48.1} \approx 1.08$.
Divide (2) by (1): $\frac{S}{S'} = \frac{1.08}{0.508} \implies \frac{S+12}{12} = 2.12 \implies S+12 = 25.4 \implies S \approx 13.4\text{ }\Omega$.
Then $R = 0.508 \times 13.4 \approx 6.8\text{ }\Omega$.
48.Why can't Wheatstone measure cell internal resistance?
Ans: A chemical cell has an active EMF. Placing it in an arm of the Wheatstone bridge introduces an additional active voltage source into the loop equations, completely breaking the simple $P/Q = R/S$ passive resistance ratio condition. The galvanometer would deflect due to the cell's EMF.
49.Procedure and derivation for internal resistance.
Ans: Step 1: Keep Key K2 open. Balance cell EMF $E$ on wire. $E = kl_1$.
Step 2: Close K2, introducing R. The cell now supplies current $I = E/(R+r)$. The potentiometer balances the terminal voltage $V = IR$. $V = kl_2$.
Divide: $E/V = l_1/l_2$.
Substitute $V = E - Ir \implies E/V = \frac{I(R+r)}{IR} = \frac{R+r}{R} = 1 + r/R$.
Equating: $1 + r/R = l_1/l_2 \implies r = R(\frac{l_1}{l_2} - 1)$.
50.Why is potentiometer an "ideal" voltmeter?
Ans: 1. It operates on a null deflection method, drawing absolutely zero current from the source being measured, thus measuring true EMF without $Ir$ drops. 2. Its sensitivity can be increased indefinitely by increasing the wire length, making it highly precise.
51.Rheostat resistance increased, what happens to null point?
Ans: Increasing rheostat resistance decreases the current in the primary circuit. This decreases the potential gradient $k$ ($V/l$). Since $E = kl$ remains constant for the experimental cell, a smaller $k$ requires a correspondingly **larger** balancing length $l$. The null point shifts to the right.
52.Driver EMF vs Experimental EMF condition.
Ans: Essential condition: $E_{driver} > E_{experimental}$. If violated ($E_{driver} < E_{experimental}$), the maximum potential drop across the entire length of the potentiometer wire will be less than the EMF to be measured. Thus, no balance point can be found anywhere on the wire.
53.Cells assist ($8\text{m}$) and oppose ($2\text{m}$). Compare EMFs.
Ans: Assisting: $E_1 + E_2 = k(8)$. Opposing: $E_1 - E_2 = k(2)$.
Divide: $\frac{E_1+E_2}{E_1-E_2} = \frac{8}{2} = 4$.
$E_1 + E_2 = 4E_1 - 4E_2 \implies 3E_1 = 5E_2 \implies \frac{E_1}{E_2} = \frac{5}{3}$.
54.Define sensitivity. How to increase?
Ans: Sensitivity is the ability to measure very small potential differences accurately, governed by the potential gradient $k$. Smaller $k$ means higher sensitivity. Increase it by: 1. Increasing the total length of the potentiometer wire. 2. Reducing the primary current by increasing the series rheostat resistance.
55.Derive shunt resistance $S$ for ammeter up to $I_0$.
Ans: Galvanometer ($G$) and Shunt ($S$) are parallel. Current $I_0$ splits: $I_g$ through $G$, $(I_0 - I_g)$ through $S$.
Parallel potential equal: $I_g \cdot G = (I_0 - I_g) \cdot S \implies S = \frac{I_g \cdot G}{I_0 - I_g}$.
56.Effective resistance of ammeter. Why low?
Ans: Parallel combination: $R_{eff} = \frac{G \cdot S}{G + S}$. Since $S$ is very small, $R_{eff} \approx S$ (very low). An ammeter is placed in series. If it had high resistance, it would significantly alter (reduce) the actual current flowing in the circuit it is trying to measure.
57.$G=12.0\text{ }\Omega, I_g=2.5\text{ mA}$. Convert to $7.5\text{ V}$ voltmeter.
Ans: Series resistance $R = \frac{V}{I_g} - G = \frac{7.5}{2.5 \times 10^{-3}} - 12.0 = \frac{7500}{2.5} - 12 = 3000 - 12 = 2988\text{ }\Omega$.
58.Effective resistance of voltmeter from Q57. Why parallel?
Ans: $R_{eff} = G + R = 12 + 2988 = 3000\text{ }\Omega$. A voltmeter measures potential difference across two points, so it must be in parallel. Its resistance must be very high so it draws negligible current from the main circuit, preventing voltage drop errors.
59.$G=50\text{ }\Omega$. Shunt for $1\%$ main current through $G$?
Ans: $I_g = 1\%\text{ of } I = 0.01 I$.
$S = \frac{I_g G}{I - I_g} = \frac{0.01 I \times 50}{I - 0.01 I} = \frac{0.01 I \times 50}{0.99 I} = \frac{50}{99} \approx 0.505\text{ }\Omega$.
60.Ammeter accidentally in parallel across high-voltage battery. Result?
Ans: An ammeter has an extremely low resistance. Connecting it directly in parallel across a high-voltage battery creates a near short-circuit. An enormously high current ($I = V / R_{ammeter}$) will flow through it, producing massive Joule heating ($I^2Rt$) that will instantly burn out and destroy the ammeter coil.
