10 questions · 1 mark each · Correct answer highlighted with explanation
10 questions · 3 marks each · Full model answers included
Ohm's law states that, at constant physical conditions (particularly temperature), the current flowing through a conductor is directly proportional to the potential difference applied across it: $V = IR$.
Limitations: Ohm's law does not hold for non-ohmic devices (like diodes, transistors, or gas discharge tubes) whose $V$-$I$ relationship is non-linear; it also fails to hold if temperature or other physical conditions change significantly with current, since resistance itself may then vary.
Resistivity ($\rho$) is an intrinsic material property describing how strongly a material opposes current flow, independent of the conductor's specific dimensions.
Conductivity ($\sigma$) is the reciprocal of resistivity: $\sigma = 1/\rho$.
Relation: $R = \rho\dfrac{l}{A}$, where $l$ is the conductor's length and $A$ is its cross-sectional area — resistance depends on both the material (via $\rho$) and the conductor's specific geometry.
Kirchhoff's current law (KCL): The algebraic sum of currents meeting at any junction in a circuit is zero — total current entering a junction equals total current leaving it (conservation of charge).
Kirchhoff's voltage law (KVL): The algebraic sum of potential differences (voltage changes) around any closed loop in a circuit is zero (conservation of energy).
In parallel, all resistors share the same voltage $V$, while total current splits among them: $I = I_1+I_2+I_3+\cdots$
Since $I_i = V/R_i$: $\dfrac{V}{R_{eq}} = \dfrac{V}{R_1}+\dfrac{V}{R_2}+\dfrac{V}{R_3}+\cdots$
Dividing by $V$: $\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\cdots$
EMF (electromotive force) is the maximum potential difference a cell can provide, measured when no current is drawn (open circuit) — it represents the total energy supplied per unit charge by the cell's internal chemical processes.
Terminal voltage is the actual potential difference across the cell's terminals when current $I$ flows, reduced from EMF due to the voltage drop across the cell's own internal resistance $r$: $V = \varepsilon - Ir$. Terminal voltage always equals EMF only when $I=0$.
Drift velocity is the average (net) velocity that free electrons acquire in a conductor due to an applied electric field, superimposed on their much larger random thermal motion.
Consider a conductor of cross-sectional area $A$, with $n$ free electrons per unit volume, each of charge $e$, drifting at speed $v_d$. In time $\Delta t$, electrons within a length $v_d\Delta t$ pass a given cross-section: charge passed $= neAv_d\Delta t$.
Current: $I = \dfrac{\text{charge}}{\text{time}} = neAv_d$
A Wheatstone bridge is a circuit of four resistors arranged in a diamond configuration, with a galvanometer connected across one diagonal and a battery across the other, used to accurately measure an unknown resistance.
The bridge is said to be "balanced" when no current flows through the galvanometer — this occurs when the four resistances satisfy the balance condition: $\dfrac{P}{Q} = \dfrac{R}{S}$, where $P,Q,R,S$ are the four arm resistances — allowing an unknown resistance to be determined precisely from three known resistances.
An ammeter measures current and is connected in series in the circuit — an ideal ammeter has zero internal resistance, so it doesn't alter the current it's measuring.
A voltmeter measures potential difference and is connected in parallel across the component of interest — an ideal voltmeter has infinite internal resistance, so it draws negligible current and doesn't alter the voltage it's measuring.
In a semiconductor, increasing temperature provides more thermal energy, exciting significantly more electrons from the valence band into the conduction band — this substantially increases the number density of free charge carriers available to conduct current.
This effect (dramatically increased carrier density) dominates over any increase in collision frequency, causing semiconductor resistivity to decrease with increasing temperature — the opposite of the behaviour seen in metals, where carrier density is already essentially fixed, and only collision frequency (which increases resistivity) changes with temperature.
Power delivered to a resistor equals the rate at which electrical energy is converted (dissipated as heat): $P = \dfrac{dW}{dt} = \dfrac{V\,dq}{dt} = VI$ (since $dq/dt = I$)
Using Ohm's law $V=IR$: $P = (IR)I = I^2R$
Equivalently, using $I=V/R$: $P = V\left(\dfrac{V}{R}\right) = \dfrac{V^2}{R}$
10 questions · 5 marks each · Detailed step-by-step solutions
(a) Equivalent resistance: $R_{eq} = 2+3+5 = 10\,\Omega$
(b) Current: $I = \dfrac{V}{R_{eq}} = \dfrac{20}{10} = \mathbf{2\text{ A}}$
(a) Equivalent resistance: $\dfrac{1}{R_{eq}} = \dfrac{1}{4}+\dfrac{1}{6} = \dfrac{3}{12}+\dfrac{2}{12} = \dfrac{5}{12}$; $R_{eq} = \dfrac{12}{5} = 2.4\,\Omega$
(b) Current division: Since voltage is the same across both, current is inversely proportional to resistance.
$I_4 = 5\times\dfrac{6}{4+6} = 5\times0.6 = 3\text{ A}$; $I_6 = 5\times\dfrac{4}{4+6} = 5\times0.4 = \mathbf{2\text{ A}}$ (check: $3+2=5$ ✓)
$R = \rho\dfrac{l}{A} = \dfrac{1.7\times10^{-8}\times2}{1\times10^{-6}}$
$R = \dfrac{3.4\times10^{-8}}{1\times10^{-6}} = \mathbf{0.034\,\Omega}$
(a) Current: $I = \dfrac{\varepsilon}{R+r} = \dfrac{6}{5.5+0.5} = \dfrac{6}{6} = 1\text{ A}$
(b) Terminal voltage: $V = \varepsilon - Ir = 6 - (1)(0.5) = \mathbf{5.5\text{ V}}$
(Verification: $V = IR_{ext} = 1\times5.5 = 5.5\text{ V}$ ✓)
(a) Current: $I = \dfrac{V}{R} = \dfrac{220}{40} = 5.5\text{ A}$
(b) Power: $P = VI = 220\times5.5 = \mathbf{1210\text{ W} = 1.21\text{ kW}}$
Volume is constant: $A_1l_1 = A_2l_2$. With $l_2 = 2l_1$: $A_2 = A_1l_1/l_2 = A_1/2$
Original: $R_1 = \rho l_1/A_1 = 10\,\Omega$. New: $R_2 = \rho l_2/A_2 = \rho(2l_1)/(A_1/2) = 4\times(\rho l_1/A_1) = 4R_1$
$R_2 = 4\times10 = \mathbf{40\,\Omega}$
Parallel combination: $\dfrac{1}{R_p} = \dfrac{1}{4}+\dfrac{1}{12} = \dfrac{3}{12}+\dfrac{1}{12} = \dfrac{4}{12} = \dfrac{1}{3}$; $R_p = 3\,\Omega$
Total resistance (series with 6 Ω): $R_{total} = 6+3 = 9\,\Omega$
Current from battery: $I = \dfrac{V}{R_{total}} = \dfrac{15}{9} \approx \mathbf{1.67\text{ A}}$
The shunt must carry the remaining current: $I_s = I - I_g = 1 - 0.001 = 0.999\text{ A}$
Since galvanometer and shunt are in parallel, they have the same voltage: $I_gG = I_sS$
$S = \dfrac{I_gG}{I_s} = \dfrac{0.001\times100}{0.999} = \dfrac{0.1}{0.999}$
$S \approx \mathbf{0.1\,\Omega}$ (approximately, since $I_s \approx I$ for a sensitive galvanometer)
Energy used per day = Power × time = $2000\text{ W} \times 5\text{ h} = 10000\text{ Wh} = 10\text{ kWh}$
Energy used in 30 days = $10\times30 = 300\text{ kWh}$ (300 units)
Total cost = $300\times6 = \mathbf{₹1800}$
Balance condition: $\dfrac{P}{Q} = \dfrac{R}{S}$
$\dfrac{10}{20} = \dfrac{15}{S}$
$S = 15\times\dfrac{20}{10} = 15\times2 = \mathbf{30\,\Omega}$