CHAPTER 3: CURRENT ELECTRICITY
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
SECTION A (16 Marks)
Q.1. The resistances of two wires having same length and same area of cross-section are $\displaystyle {2\,\Omega}$ and $\displaystyle {8\,\Omega}$ respectively. The ratio of their specific resistances (resistivities) is:
Explanation: $\displaystyle {R = \rho {\frac{{l}}{{A}}} \implies \rho \propto R}$. Since length $l$ and area $A$ are identical, $\displaystyle {{\frac{{\rho_1}}{{\rho_2}}} = {\frac{{R_1}}{{R_2}}} = {\frac{{2}}{{8}}} = {\frac{{1}}{{4}}} = 1:4}$.
Q.2. Kirchhoff’s junction rule is a reflection of the law of conservation of:
Explanation: The total charge entering a junction in an electric circuit per unit time equals the total charge leaving it per unit time ($\displaystyle {\sum I = 0}$).
Q.3. Which of the following material has a negative temperature coefficient of resistance ($\displaystyle {\alpha < 0}$)?
Explanation: Germanium is a semiconductor. Its resistivity decreases exponentially with an increase in temperature as more charge carriers are thermally generated.
Q.4. A wire of resistance $\displaystyle {R}$ is stretched to double its original length keeping volume constant. Its new resistance will be:
Explanation: Volume $\displaystyle {V = A \cdot l}$ is constant. When length becomes $\displaystyle {l' = 2l}$, area becomes $\displaystyle {A' = A/2}$. $\displaystyle {R' = \rho {\frac{{l'}}{{A'}}} = \rho {\frac{{2l}}{{A/2}}} = 4 \rho {\frac{{l}}{{A}}} = 4R}$.
Q.5. The electric drift velocity $\displaystyle {v_d}$ of conduction electrons in a metal conductor is related to applied electric field $\displaystyle {E}$ as:
Explanation: Drift velocity is given by $\displaystyle {v_d = e E \tau / m = \mu E}$, where $\displaystyle {\mu}$ is mobility.
Q.6. Two cells of EMFs $\displaystyle {E_1}$ and $\displaystyle {E_2}$ and internal resistances $\displaystyle {r_1}$ and $\displaystyle {r_2}$ are connected in parallel. Equivalent internal resistance of combination is:
Explanation: Internal resistances act as parallel resistors, so $\displaystyle {{\frac{{1}}{{r_{\text{eq}}}}} = {\frac{{1}}{{r_1}}} + {\frac{{1}}{{r_2}}} \implies r_{\text{eq}} = {\frac{{r_1 r_2}}{{r_1 + r_2}}}}$.
Q.7. When a current of $\displaystyle {0.2\text{ A}}$ is drawn from a battery, terminal potential difference is $\displaystyle {20\text{ V}}$. When a current of $\displaystyle {2\text{ A}}$ is drawn, terminal voltage drops to $\displaystyle {16\text{ V}}$. EMF of battery is:
Explanation: $\displaystyle {V = E - I r}$. Equation 1: $\displaystyle {20 = E - 0.2 r}$. Equation 2: $\displaystyle {16 = E - 2.0 r}$. Subtracting gives $\displaystyle {4 = 1.8 r \implies r = 2.22\,\Omega}$. Thus $\displaystyle {E = 20 + 0.2(2.22) = 20.44\text{ V} \approx 20.4\text{ V}}$.
Q.8. What happens to the resistance of a metallic conductor when its temperature increases?
Explanation: Higher temperature causes metallic lattice ions to vibrate more violently, increasing electron collision frequency and decreasing relaxation time $\displaystyle {\tau}$, which increases resistance $\displaystyle {R \propto 1/\tau}$.
Q.9. A battery of EMF $\displaystyle {12\text{ V}}$ and internal resistance $\displaystyle {1\,\Omega}$ is connected to a $\displaystyle {5\,\Omega}$ resistor. The terminal voltage across battery is:
Explanation: Circuit current $\displaystyle {I = {\frac{{E}}{{R + r}}} = {\frac{{12}}{{5 + 1}}} = 2\text{ A}}$. Terminal voltage $\displaystyle {V = I R = 2 \times 5 = 10\text{ V}}$ (or $\displaystyle {V = E - I r = 12 - (2)(1) = 10\text{ V}}$).
Q.10. Mobility of free electrons in a conductor is defined as:
Explanation: Mobility $\displaystyle {\mu = {\frac{{v_d}}{{E}}} = {\frac{{e \tau}}{{m}}}}$. SI Unit: $\displaystyle {\text{m}^2\text{V}^{-1}\text{s}^{-1}}$.
Q.11. If the percentage change in current flowing through a fixed resistor is $1\%$, the percentage change in electric power dissipated will be approximately:
Explanation: $\displaystyle {P = I^2 R}$. By differentiation / error approximation, $\displaystyle {{\frac{{\Delta P}}{{P}}} \times 100\% = 2 {\left( {\frac{{\Delta I}}{{I}}} \right)} \times 100\% = 2 \times 1\% = 2\%}$.
Q.12. The voltage-current ($\displaystyle {V-I}$) graph for an Ohmic conductor is a:
Explanation: According to Ohm's law, $\displaystyle {V \propto I}$, so the graph of $\displaystyle {V}$ vs $\displaystyle {I}$ is a straight line whose slope represents constant resistance $\displaystyle {R = V/I}$.
Directions for Q.13 to Q.16:
(A) Both Assertion and Reason are true and Reason is correct explanation of Assertion.
(B) Both Assertion and Reason are true but Reason is NOT correct explanation.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.
Q.13. Assertion (A): A voltmeter connected across the terminals of a cell always gives the exact EMF of the cell.
Reason (R): Terminal potential difference of a cell drawing current $i$ is $\displaystyle {V = E - i r}$.
Explanation: A real voltmeter draws some current, so it measures terminal voltage $\displaystyle {V = E - i r < E}$, not EMF. EMF is measured only when no current flows ($i=0$).
Q.14. Assertion (A): A domestic electrical appliance working on a three-pin plug will continue working even if the top earth pin is removed.
Reason (R): The third earth pin is used purely as a safety grounding device.
Explanation: Current flows through live and neutral pins; earth pin protects the user against leakage shocks.
Q.15. Assertion (A): Insulators do not allow flow of electric current through them.
Reason (R): Insulators have almost no free charge carriers available for conduction.
Q.16. Assertion (A): When length of a conductor is doubled without changing cross-sectional area, its resistance doubles.
Reason (R): Resistance of a conductor is directly proportional to its length ($\displaystyle {R \propto l}$).
SECTION B (10 Marks)
Q.17. The following graph shows potential difference $\displaystyle {V}$ across terminals of a cell against load current $\displaystyle {I}$.
Find (i) EMF of the cell, and (ii) Internal resistance of the cell.
(i) EMF $\displaystyle {E}$ equals y-intercept when $\displaystyle {I = 0 \implies E = 2.0\text{ V}}$.
(ii) Short circuit current when $\displaystyle {V = 0 \implies I_{\text{sc}} = 4.0\text{ A}}$. Internal resistance $\displaystyle {r = {\frac{{E}}{{I_{\text{sc}}}}} = {\frac{{2.0}}{{4.0}}} = 0.5\,\Omega}$.
Q.18. $N$ identical resistors each of resistance $R$, when connected in series have effective resistance $X$, and when connected in parallel have effective resistance $Y$. Find relation between $R$, $X$, and $Y$.
Series: $\displaystyle {X = N R \implies R = {\frac{{X}}{{N}}}}$.
Parallel: $\displaystyle {Y = {\frac{{R}}{{N}}} \implies R = N Y}$.
Multiplying both: $\displaystyle {X Y = (N R) {\left( {\frac{{R}}{{N}}} \right)} = R^2 \implies R = \sqrt{X Y}}$.
Q.19. Resistance of a conductor is given by $\displaystyle {R = {\frac{{m l}}{{n e^2 \tau A}}}}$. State how $R$ varies with (i) length $l$, (ii) relaxation time $\displaystyle {\tau}$.
(i) $\displaystyle {R \propto l}$ (Resistance is directly proportional to length).
(ii) $\displaystyle {R \propto 1/\tau}$ (Resistance is inversely proportional to relaxation time $\displaystyle {\tau}$).
Q.20(I). Graph for a metallic wire at two different temperatures $\displaystyle {T_1}$ and $\displaystyle {T_2}$ is shown in figure. Which temperature is higher and why?
Q.20(II). A car battery has EMF $\displaystyle {12\text{ V}}$ and internal resistance $\displaystyle {0.4\,\Omega}$. What is the maximum current that can be drawn from the battery?
Part I: Slope of $V-I$ graph gives resistance $R$. The line with steeper slope has higher resistance. Since metallic resistance increases with temperature, higher slope corresponds to higher temperature ($\displaystyle {T_1 > T_2}$).
OR Part II: Maximum current occurs on short circuit ($\displaystyle {R = 0}$): $\displaystyle {I_{\text{max}} = {\frac{{E}}{{r}}} = {\frac{{12}}{{0.4}}} = 30\text{ A}}$.
Q.21(I). Potential difference applied across a given conductor is doubled. How will this affect (i) mobility of electrons, (ii) current density in conductor? Justify.
Q.21(II). Two metallic wires of same material have same length but cross-sectional areas in ratio $1:2$. Compare drift velocities of electrons when connected (i) in series, (ii) in parallel.
Part I: (i) Mobility $\displaystyle {\mu = e \tau / m}$ is independent of applied potential $V$, so it remains unchanged.
(ii) Current density $\displaystyle {J = \sigma E = \sigma V / l}$. When $V$ is doubled, electric field $E$ doubles, so current density $J$ doubles.
OR Part II: (i) In series, current $I$ is same: $\displaystyle {I = n e A v_d \implies v_d \propto 1/A \implies {\frac{{v_{d1}}}{{v_{d2}}} = {\frac{{A_2}}{{A_1}}} = {\frac{{2}}{{1}}} = 2:1}}$.
(ii) In parallel, potential $V$ and electric field $E = V/l$ are same: $\displaystyle {v_d = e E \tau / m}$, independent of area, so $\displaystyle {v_{d1} : v_{d2} = 1:1}$.
SECTION C (21 Marks)
Q.22. A battery of EMF $\displaystyle {6\text{ V}}$ and internal resistance $\displaystyle {2\,\Omega}$ is connected to a resistor. If current in circuit is $\displaystyle {0.25\text{ A}}$, find (i) resistance of resistor, (ii) terminal voltage of battery.
(i) $\displaystyle {I = {\frac{{E}}{{R + r}}} \implies 0.25 = {\frac{{6}}{{R + 2}}} \implies R + 2 = 24 \implies R = 22\,\Omega}$.
(ii) Terminal voltage $\displaystyle {V = I R = 0.25 \times 22 = 5.5\text{ V}}$ (or $\displaystyle {V = E - I r = 6 - (0.25)(2) = 5.5\text{ V}}$).
Q.23(I). (A) State the fundamental conservation law behind Kirchhoff’s junction rule ($\displaystyle {I_1 = I_2 + I_3}$).
(B) Explain the validation of law of conservation of energy in Kirchhoff's loop rule.
Q.23(II). How is the balancing condition affected if galvanometer and cell are interchanged in a Wheatstone bridge?
(A) Law of Conservation of Electric Charge.
(B) Electrostatic force is conservative. Total work done in taking a unit charge around any closed loop is zero ($\displaystyle {\sum \Delta V = 0}$).
OR Part: The null balance condition ($\displaystyle {P/Q = R/S}$) remains completely unchanged; galvanometer will still show zero deflection.
Q.24. Network PQRS has batteries $5\text{ V}$ and $10\text{ V}$ with negligible internal resistance. Milliammeter of $50\,\Omega$ resistance is connected between P and R. Calculate reading of milliammeter.
Calculated current through milliammeter branch $\displaystyle {I = 0.04\text{ A} = 40\text{ mA}}$.
Q.25. Use Kirchhoff’s rules to determine value of current $\displaystyle {I_1}$ flowing in circuit shown in figure.
Loop 1 equation: $\displaystyle {30 I_1 + 20 I_3 = 20}$.
Loop 2 equation: $\displaystyle {20 I_2 + 20 I_3 = 80 \implies I_2 + I_3 = 4}$.
Solving system of simultaneous equations gives $\displaystyle {I_1 = 4\text{ A}}$.
Q.26. Uniform wire of resistance $12\,\Omega$ is cut into 3 pieces with ratio $R_1 : R_2 : R_3 = 1 : 2 : 3$. Connected to form a triangle across which cell of EMF $8\text{ V}$ and internal resistance $1\,\Omega$ is connected. Calculate current through each part.
Triangle sides form parallel branches: branch 1 is $R_3 = 6\,\Omega$, branch 2 is $R_1 + R_2 = 2 + 4 = 6\,\Omega$.
Parallel combination $\displaystyle {R_p = {\frac{{6 \times 6}}{{6 + 6}}} = 3\,\Omega}$. Total resistance $\displaystyle {R_{\text{total}} = 3 + 1 = 4\,\Omega}$.
Total cell current $\displaystyle {I = {\frac{{8}}{{4}}} = 2\text{ A}}$. Current splits equally as $1\text{ A}$ through $R_3$ and $1\text{ A}$ through $R_1$ and $R_2$.
Q.27(I). State Kirchhoff’s rules. Explain briefly how these rules are justified.
Q.27(II). Estimate average drift speed of conduction electrons in copper wire of cross-section $2.5 \times 10^{-7}\text{ m}^2$ carrying current $1.8\text{ A}$. ($\displaystyle {n = 9 \times 10^{28}\text{ m}^{-3}}$).
Part I: Junction Rule: $\displaystyle {\sum I = 0}$ (charge conservation). Loop Rule: $\displaystyle {\sum \Delta V = 0}$ (energy conservation).
OR Part II: $\displaystyle {v_d = {\frac{{I}}{{n e A}}} = {\frac{{1.8}}{{(9 \times 10^{28}) (1.6 \times 10^{-19}) (2.5 \times 10^{-7})}}} = 5 \times 10^{-4}\text{ m/s} = 0.5\text{ mm/s}}$.
Q.28. Draw a plot showing variation of resistivity $\displaystyle {\rho}$ with temperature $\displaystyle {T}$ for (i) a conductor (copper), (ii) a semiconductor (silicon).
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: Temperature Dependence of Resistance.
Variation of resistance with temperature is given by $\displaystyle {R_T = R_0 (1 + \alpha \Delta T)}$, where $\displaystyle {\alpha}$ is temperature coefficient of resistance.
(i) For metals, $\displaystyle {\alpha}$ is: (a) Positive (b) Negative (c) Zero
(ii) Material used for standard resistance coil: (a) Copper (b) Manganin (c) Aluminium
(iii) Resistance of wire at $0^\circ\text{C}$ is $10\,\Omega$ and at $100^\circ\text{C}$ is $12\,\Omega$. Find $\displaystyle {\alpha}$: (a) $0.002/^\circ\text{C}$ (b) $0.02/^\circ\text{C}$
Q.30. Case Study 2: Internal Resistance and Terminal Potential Difference.
Terminal voltage $\displaystyle {V = E - I r}$ during discharging and $\displaystyle {V = E + I r}$ during charging.
(i) Current drawn when battery is short-circuited: (a) $\displaystyle {E/r}$ (b) Zero (c) Infinite
(ii) Terminal voltage equals EMF when: (a) Circuit is open ($I=0$) (b) Short circuited
(iii) Internal resistance of cell increases with: (a) Increase in concentration of electrolyte (b) Decrease in distance between plates
SECTION E (15 Marks)
Q.31(I). (A) Derive relation between current density $\displaystyle {\vec{J}}$ and potential difference $\displaystyle {V}$ across a conductor: $\displaystyle {J = \sigma E = \sigma V / l}$.
(B) Estimate drift speed of electrons in copper wire of area $1.0 \times 10^{-7}\text{ m}^2$ carrying current $1.5\text{ A}$ ($\displaystyle {n = 9 \times 10^{28}\text{ m}^{-3}}$).
Q.31(II). (A) Why do free electrons flowing by themselves in metal wire not cause net current?
(B) Derive Ohm's law vector form $\displaystyle {\vec{J} = \sigma \vec{E}}$ from electron drift concept.
Part I (B): $\displaystyle {v_d = {\frac{{I}}{{n e A}}} = {\frac{{1.5}}{{(9 \times 10^{28})(1.6 \times 10^{-19})(1.0 \times 10^{-7})}}} = 1.04 \times 10^{-3}\text{ m/s} = 1.04\text{ mm/s}}$.
OR Part II (A): Random thermal motion in all directions averages to zero net velocity ($\displaystyle {\sum \vec{u} = 0}$), hence zero net current without applied electric field.
Q.32(I). (a) What is Wheatstone bridge?
(b) Derive condition for bridge to be balanced ($\displaystyle {P/Q = R/S}$).
(c) Resistance $10\,\Omega$ in left gap and unknown $X$ in right gap of meter bridge. Balance point at $40\text{ cm}$. Calculate $X$.
Q.32(II). (a) State Ohm’s law.
(b) Derive expression for current using Ohm's law ($\displaystyle {I = V / R}$).
(c) Define resistance and factors affecting it.
(d) Draw V-I graphs for (i) Ohmic conductor, (ii) Non-Ohmic conductor (p-n junction diode).
Part I (c): Meter bridge formula: $\displaystyle {{\frac{{10}}{{X}}} = {\frac{{l}}{{100 - l}}} = {\frac{{40}}{{60}}} \implies X = {\frac{{10 \times 60}}{{40}}} = 15\,\Omega}$.
Q.33(I). Explain variation of resistance with temperature in (a) Conductors, (b) Semiconductors, (c) Insulators with neat labelled graphs.
Q.33(II). Two cells of EMFs $\displaystyle {E_1, E_2}$ and internal resistances $\displaystyle {r_1, r_2}$ connected in (a) Series, (b) Parallel. Derive expressions for net EMF and effective internal resistance in both cases.
Series: $\displaystyle {E_{\text{eq}} = E_1 + E_2}$, $\displaystyle {r_{\text{eq}} = r_1 + r_2}$.
Parallel: $\displaystyle {E_{\text{eq}} = {\frac{{E_1 r_2 + E_2 r_1}}{{r_1 + r_2}}}}$, $\displaystyle {r_{\text{eq}} = {\frac{{r_1 r_2}}{{r_1 + r_2}}}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. The SI unit of resistivity is:
2. Two wires of same material and length have radii in ratio $2:1$. Ratio of their resistances is:
Explanation: $\displaystyle {R = \rho {\frac{{l}}{{\pi r^2}}} \propto {\frac{{1}}{{r^2}}} \implies {\frac{{R_1}}{{R_2}}} = {\left( {\frac{{r_2}}{{r_1}}} \right)}^2 = {\left( {\frac{{1}}{{2}}} \right)}^2 = {\frac{{1}}{{4}}}}$.
3. Kirchhoff’s First Law ($\displaystyle {\sum I = 0}$) is based on conservation of:
4. A cell of EMF $1.5\text{ V}$ and internal resistance $1\,\Omega$ is connected to external resistor $2\,\Omega$. Current in circuit is:
Explanation: $\displaystyle {I = {\frac{{E}}{{R + r}}} = {\frac{{1.5}}{{2 + 1}}} = 0.5\text{ A}}$.
5. A wire of resistance $R$ is stretched to double its length keeping volume constant. New resistance is:
6. Drift velocity of electrons increases when:
7. A wire has resistance $R$. Cut into 3 equal parts and connected in parallel. New resistance is:
Explanation: Each piece has resistance $\displaystyle {R' = R/3}$. Three equal pieces in parallel give $\displaystyle {R_{\text{net}} = R'/3 = R/9}$.
8. When 5 cells each of EMF $2\text{ V}$ and internal resistance $0.5\,\Omega$ are connected in series, total EMF is:
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): Ohm's law is not universally applicable to all conducting elements (e.g. semiconductors).
Reason (R): The resistance of semiconductor materials changes non-linearly with applied voltage and temperature.
2. Assertion (A): The internal resistance of a cell increases gradually with prolonged usage.
Reason (R): Chemical reactions inside the electrolyte lead to deposition of reaction products on electrodes and depletion of free ions.
3. Assertion (A): Bending a current-carrying copper wire does not affect its electrical resistance.
Reason (R): Resistance depends on length, cross-sectional area, and resistivity of material, which remain unchanged on bending.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study 1: Mobile Phone Charging & Power Applications
A mobile charger converts $220\text{ V}$ AC supply into $5\text{ V}, 2\text{ A}$ DC output to safely charge a lithium-ion battery.
Q1. Why is AC converted to DC before charging?
Ans: Chemical energy storage in batteries requires unidirectional (DC) current flow.
Q2. Calculate maximum power supplied by charger:
Ans: $\displaystyle {P = V I = 5 \times 2 = 10\text{ W}}$.
Q3. Effect of using high resistance charging cable?
Ans: Causes potential drop ($\displaystyle {I R}$) and heat loss ($\displaystyle {I^2 R}$), slowing down charging speed.
Q4. Why are copper wires preferred in charger cables?
Ans: Copper has high electrical conductivity and low resistivity ($\displaystyle {\rho}$).
ASSIGNMENT – 5 (CONCEPTUAL & DERIVATIONS)
SHORT ANSWER QUESTIONS
1. A battery of $6\text{ V}$ drives $60\text{ mA}$ through an electric lamp. Another battery of $10\text{ V}$ drives $70\text{ mA}$ through same lamp. Is lamp Ohmic?
2. Two cells of EMFs $1.5\text{ V}$ and $2.0\text{ V}$ having internal resistances $0.2\,\Omega$ and $0.3\,\Omega$ connected in parallel. Calculate equivalent EMF and internal resistance.
$\displaystyle {E_{\text{eq}} = {\frac{{E_1 r_2 + E_2 r_1}}{{r_1 + r_2}}} = {\frac{{(1.5)(0.3) + (2.0)(0.2)}}{{0.5}}} = {\frac{{0.45 + 0.40}}{{0.5}}} = 1.7\text{ V}}$.
3. A wire of resistance $8R$ is bent into a complete circle. What is effective resistance across diameter AB?
$\displaystyle {R_{\text{eff}} = {\frac{{4R \times 4R}}{{4R + 4R}}} = 2R}$.
LONG ANSWER QUESTIONS / DERIVATIONS
1. Derive expression for electric current in terms of drift velocity: $\displaystyle {I = n e A v_d}$.
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