NEET Physics: Electrostatics and Current Electricity · Chapter 3

Current Electricity

What to remember

Ohm's law: V = IR for a conductor at constant temperature. Resistance R = ρl/A, and current I = neAv_d, where v_d is the drift velocity.
A cell of emf ε and internal resistance r gives terminal voltage V = ε – Ir while delivering current, and V = ε + Ir while being charged. Open circuit: V = ε.
Kirchhoff's rules: the sum of currents at a junction is zero (charge conservation), and the sum of potential changes around a closed loop is zero (energy conservation).
Wheatstone bridge balance: P/Q = R/S, so no current flows through the galvanometer. Metre bridge: unknown X = R(100 – l)/l.

Electric current, flow of charge and drift velocity

Electric current is the rate of flow of charge through a cross-section: I = dq/dt. Its SI unit is the ampere (A) = C s⁻¹. By convention, current flows from the positive terminal to the negative terminal outside the cell, which is the direction opposite to electron flow. Current is a scalar, although it has a direction along the wire.

In a metal, free electrons move randomly at very high speeds (of the order of 10⁶ m s⁻¹) and collide with the fixed ions. With no field the average velocity is zero. When an electric field E is applied, each electron gets an acceleration a = eE/m between collisions. Over the average time τ between collisions (relaxation time) the electrons acquire a small average velocity opposite to the field, the drift velocity:

v_d = (eE/m) τ

The drift speed is very small, of the order of 10⁻⁴ m s⁻¹, yet a bulb lights almost instantly because the field is set up along the whole wire at nearly the speed of light.

Mobility μ = v_d/E = eτ/m. Its unit is m² V⁻¹ s⁻¹.

Relation between current and drift velocity: I = neAv_d. Here n is the number of free electrons per unit volume (m⁻³), A is the area of cross-section (m²), e = 1.6 × 10⁻¹⁹ C. Current density J = I/A = nev_d, in A m⁻².

Ohm's law, resistance, resistivity and conductivity

Ohm's law: at constant temperature, the current through a conductor is proportional to the potential difference across it: V = IR. The constant R is the resistance, with unit ohm (Ω) = V A⁻¹.

For a uniform conductor of length l and cross-section A:

R = ρl/A

where ρ is the resistivity of the material. Resistivity depends on the material and temperature, not on the shape or size. Its unit is Ω m. Conductivity σ = 1/ρ, in Ω⁻¹ m⁻¹ (S m⁻¹). The microscopic form of Ohm's law is J = σE.

From the drift model, ρ = m/(ne²τ). So resistivity falls if n or τ is larger.

Quantity	Symbol	Depends on	Unit
Resistance	R	Material, length, area, temperature	Ω
Resistivity	ρ	Material, temperature	Ω m
Conductance	G = 1/R	Same as R	S
Conductivity	σ = 1/ρ	Material, temperature	S m⁻¹

Stretching a wire to n times its length at constant volume makes the area 1/n times, so R becomes n² times. Cutting a wire into n equal parts makes each part R/n.

V-I characteristics

Ohmic conductors (metals at constant temperature): the V–I graph is a straight line through the origin; the slope of I against V is 1/R.
Non-ohmic devices (diode, thermistor, filament lamp, electrolytes): the V–I graph is curved or non-linear. A p–n junction diode has an asymmetric graph: current is very small in reverse bias and rises sharply in forward bias beyond a threshold voltage.
For a bulb filament the graph bends towards the voltage axis because resistance increases as the filament heats up.

Temperature dependence of resistance and resistivity

For a metallic conductor over a modest temperature range:

ρ_T = ρ₀[1 + α(T – T₀)], R_T = R₀[1 + α(T – T₀)]

Here α is the temperature coefficient of resistivity, in K⁻¹ (or °C⁻¹).

Material type	Effect of rising temperature	Reason	α
Metals (Cu, Ag)	Resistivity increases	More frequent collisions; τ decreases	Positive
Semiconductors (Si, Ge)	Resistivity decreases	Number density n of carriers rises sharply	Negative
Insulators	Resistivity decreases	More carriers freed	Negative
Alloys (nichrome, manganin, constantan)	Resistivity changes very little	Very small α	Small positive

Manganin and constantan are used for standard resistors and wire-wound resistors because of their small α and high resistivity. Nichrome is used in heating elements because of its high resistivity and high melting point.

Electrical energy and power

When a charge q moves through a potential difference V, the work done is qV. The power delivered is

P = VI = I²R = V²/R (unit: watt, W)

Energy in time t is E = Pt = I²Rt (joule heating, Joule's law). The commercial unit is the kilowatt-hour: 1 kWh = 3.6 × 10⁶ J.

For a fixed supply voltage, a bulb with a smaller resistance has more power (P = V²/R). For the same current, a larger resistance dissipates more power (P = I²R). Two bulbs in series share the same current, so the one with higher resistance glows brighter. Two bulbs in parallel share the same voltage, so the one with lower resistance glows brighter.

Combination of resistors

Arrangement	Equivalent	Current	Voltage
Series	R = R₁ + R₂ + ...	Same in all	Divides in proportion to R
Parallel	1/R = 1/R₁ + 1/R₂ + ...	Divides in inverse proportion to R	Same across all

Two resistors in parallel: R = R₁R₂/(R₁ + R₂). The equivalent in parallel is less than the smallest resistance. For n equal resistors of value R, series gives nR and parallel gives R/n.

Cells, emf and internal resistance

The emf ε of a cell is the potential difference between its terminals when no current is drawn (open circuit). It is the work done by the cell per unit charge to move charge round the full circuit. Every cell has internal resistance r.

When the cell supplies current I to an external resistance R:

I = ε/(R + r), terminal voltage V = ε – Ir = IR

When the cell is being charged by a larger source, V = ε + Ir. When the circuit is shorted (R = 0) the current is ε/r, the maximum possible. The terminal voltage is less than ε in discharge, and V = ε when I = 0.

Cells in series (n identical cells, each ε and r): total emf = nε and total internal resistance = nr. Current I = nε/(R + nr). Cells connected with opposite polarity subtract.

Cells in parallel (n identical cells): total emf = ε and total internal resistance = r/n. Current I = ε/(R + r/n).

For m rows each of n cells in series, connected in parallel: ε_eq = nε, r_eq = nr/m. The current is largest when the external resistance R equals the equivalent internal resistance (for a given total number of cells). Maximum power is delivered to the load when R = r.

Kirchhoff's rules

Junction rule: the algebraic sum of currents meeting at a junction is zero. Current entering equals current leaving. This expresses conservation of charge.

Loop rule: the algebraic sum of changes in potential around any closed loop is zero. This expresses conservation of energy.

Sign convention when going around a loop:

Step	Change in potential
Through a resistor, in the direction of current	–IR
Through a resistor, against the current	+IR
Through a cell from the negative to the positive terminal	+ε
Through a cell from the positive to the negative terminal	–ε

Write one equation per independent loop and per independent junction, then solve for the unknown currents. A negative answer means the actual direction is opposite to the assumed direction.

Wheatstone bridge

Four resistors P, Q, R and S form a quadrilateral with a cell across one diagonal and a galvanometer across the other. The bridge is balanced when the galvanometer shows zero deflection. The balance condition is

P/Q = R/S

At balance, the two junctions that the galvanometer joins are at the same potential, so the resistor between them (or the galvanometer) carries no current and can be removed from the calculation. The balance condition does not change if the cell and galvanometer are interchanged.

Metre bridge

A metre bridge is a practical Wheatstone bridge. A uniform wire of length 100 cm is stretched along a metre scale. A known resistance R is in one gap and an unknown resistance X is in the other gap. A jockey slides on the wire and finds the balance point at distance l from one end. Because the wire is uniform, resistance is proportional to length:

R/X = l/(100 – l), so X = R(100 – l)/l

Best accuracy is when the balance point is near the middle of the wire.

A potentiometer is an older instrument for comparing emfs.

Colour code of resistors

Carbon resistors have coloured rings. The first two rings give the first two digits, the third ring gives the multiplier (power of ten), and the last ring gives the tolerance.

Colour	Digit	Colour	Digit
Black	0	Green	5
Brown	1	Blue	6
Red	2	Violet	7
Orange	3	Grey	8
Yellow	4	White	9

Tolerance: gold ±5%, silver ±10%, no ring ±20%. Example: yellow, violet, red, gold = 47 × 10² Ω = 4700 Ω ± 5%.

Formula summary

Quantity	Formula
Current	I = dq/dt = neAv_d
Drift velocity	v_d = eEτ/m
Mobility	μ = v_d/E
Resistance	R = ρl/A
Temperature effect	R_T = R₀[1 + α(T – T₀)]
Power	P = VI = I²R = V²/R
Cell	V = ε – Ir
Series / parallel	R = ΣRᵢ; 1/R = Σ(1/Rᵢ)
Wheatstone	P/Q = R/S
Metre bridge	X = R(100 – l)/l

Worked examples

Example 1 (drift velocity). A wire of cross-section 10⁻⁶ m² carries 1.6 A. The free electron density is 10²⁹ m⁻³. v_d = I/(neA) = 1.6/(10²⁹ × 1.6 × 10⁻¹⁹ × 10⁻⁶) = 10⁻⁴ m s⁻¹.

Example 2 (resistance). A wire of 4 Ω is stretched uniformly to twice its length. The new resistance is 4 × 2² = 16 Ω.

Example 3 (internal resistance). A cell of emf 12 V and internal resistance 1 Ω is connected to a 5 Ω resistor. I = 12/(5 + 1) = 2 A. The terminal voltage V = 12 – 2 × 1 = 10 V = 2 × 5 V.

Example 4 (Kirchhoff's rule). Two cells in opposition: cell 1 of 10 V and cell 2 of 4 V (negligible internal resistance) joined with a total resistance of 3 Ω in a single loop. Net emf = 10 – 4 = 6 V, so I = 6/3 = 2 A.

Example 5 (metre bridge). With R = 10 Ω in the left gap, balance is at 40 cm. X = 10 × (100 – 40)/40 = 15 Ω.

Example 6 (power). A heater rated 1000 W at 200 V has resistance R = V²/P = 40000/1000 = 40 Ω. At 100 V it gives P = 10000/40 = 250 W.

NEET traps

1. Drift speed and signal speed. Drift speed is about 10⁻⁴ m s⁻¹; the effect of closing the switch travels at nearly the speed of light.
2. Drift velocity is opposite to E for electrons, while conventional current is along E.
3. Resistivity versus resistance. Resistivity is a property of the material; resistance also depends on length and area.
4. Temperature effect. Metals: R rises with T. Semiconductors: R falls with T. Do not mix them up.
5. Stretching a wire. R becomes n² times for n times the length, not n times.
6. Series versus parallel bulbs. In series the higher-resistance (lower-power rated) bulb glows brighter; in parallel the lower-resistance (higher-power rated) bulb glows brighter.
7. Terminal voltage versus emf. V is less than ε during discharge, greater than ε during charging, and equal to ε in open circuit.
8. Cells in parallel keep the same emf; only the internal resistance falls to r/n.
9. Balanced Wheatstone bridge. The galvanometer current is zero, but current flows in the other four arms.
10. Colour code. The third ring is a multiplier, not a digit. Gold is ±5% and silver is ±10%.

One-liners

1. The SI unit of current is the ampere, which equals C s⁻¹.
2. Conventional current flows opposite to the flow of electrons.
3. Drift velocity is proportional to the applied electric field.
4. Mobility has the unit m² V⁻¹ s⁻¹.
5. The unit of resistivity is Ω m.
6. Conductivity is the reciprocal of resistivity.
7. The V–I graph of an ohmic conductor is a straight line through the origin.
8. The resistivity of a semiconductor decreases as temperature increases.
9. Manganin and constantan have very small temperature coefficients.
10. 1 kWh = 3.6 × 10⁶ J.
11. A shorted cell delivers current ε/r.
12. For n identical cells in series the total emf is nε.
13. The junction rule follows from conservation of charge.
14. The loop rule follows from conservation of energy.
15. A Wheatstone bridge is balanced when P/Q = R/S.

Practice questions

A copper wire of cross-section 2 × 10⁻⁶ m² has 10²⁹ free electrons per m³. If the drift speed is 5 × 10⁻⁵ m s⁻¹, the current is (e = 1.6 × 10⁻¹⁹ C)
1. 3.2 A
2. 16 A
3. 0.16 A
4. 1.6 A
Answer
D. 1.6 A
I = neAv = 10²⁹ × 1.6 × 10⁻¹⁹ × 2 × 10⁻⁶ × 5 × 10⁻⁵ = 1.6 A.
A steady current of 3 A flows in a wire for 20 s. The charge that passes through a cross-section is
1. 60 C
2. 6.7 C
3. 23 C
4. 0.15 C
Answer
A. 60 C
q = It = 3 × 20 = 60 C.
A wire of resistance 4 Ω is stretched uniformly to three times its length. Its new resistance is
1. 18 Ω
2. 36 Ω
3. 12 Ω
4. 108 Ω
Answer
B. 36 Ω
Volume is constant, so A becomes A/3. R ∝ l²: R' = 4 × 3² = 36 Ω.
A copper wire has resistivity 1.7 × 10⁻⁸ Ω m, length 100 m and cross-section 10⁻⁶ m². Its resistance is
1. 1.7 × 10⁻⁴ Ω
2. 17 Ω
3. 1.7 Ω
4. 0.17 Ω
Answer
C. 1.7 Ω
R = ρl/A = 1.7 × 10⁻⁸ × 100/10⁻⁶ = 1.7 Ω.
The resistance of a metal wire is 100 Ω at 20 °C. If its temperature coefficient of resistance is 4 × 10⁻³ °C⁻¹, the resistance at 70 °C is
1. 104 Ω
2. 200 Ω
3. 80 Ω
4. 120 Ω
Answer
D. 120 Ω
R = 100[1 + 4 × 10⁻³ × 50] = 100 × 1.2 = 120 Ω.
A bulb is rated 100 W, 220 V. Its resistance is about
1. 484 Ω
2. 22000 Ω
3. 48.4 Ω
4. 2.2 Ω
Answer
A. 484 Ω
R = V²/P = 220²/100 = 48400/100 = 484 Ω.
The energy consumed by a 1 kW heater in 2 hours is
1. 1.2 × 10⁵ J
2. 3.6 × 10⁶ J
3. 7.2 × 10⁶ J
4. 7.2 × 10⁴ J
Answer
C. 7.2 × 10⁶ J
E = Pt = 1000 × 7200 = 7.2 × 10⁶ J (2 kWh).
Two resistors of 12 Ω and 4 Ω are connected in parallel. The equivalent resistance is
1. 0.33 Ω
2. 3 Ω
3. 16 Ω
4. 8 Ω
Answer
B. 3 Ω
R = (12 × 4)/(12 + 4) = 48/16 = 3 Ω.
A cell of emf 9 V and internal resistance 0.5 Ω is connected to a 4 Ω resistor. The terminal voltage is
1. 7 V
2. 8 V
3. 9 V
4. 4.5 V
Answer
B. 8 V
I = 9/(4 + 0.5) = 2 A. V = ε – Ir = 9 – 1 = 8 V (also IR = 2 × 4 = 8 V).
A cell of emf 1.5 V and internal resistance 0.5 Ω is short-circuited. The current is
1. 0.75 A
2. 2 A
3. 3 A
4. 1.5 A
Answer
C. 3 A
I = ε/r = 1.5/0.5 = 3 A.
In a balanced Wheatstone bridge P = 10 Ω, Q = 20 Ω and R = 30 Ω, where P/Q = R/S. The value of S is
1. 60 Ω
2. 15 Ω
3. 6 Ω
4. 45 Ω
Answer
A. 60 Ω
P/Q = R/S gives S = QR/P = 20 × 30/10 = 60 Ω.
In a metre bridge, a known resistance of 15 Ω is in the left gap and the balance point is at 60 cm from the left end. The unknown resistance in the right gap is
1. 22.5 Ω
2. 9 Ω
3. 25 Ω
4. 10 Ω
Answer
D. 10 Ω
X = R(100 – l)/l = 15 × 40/60 = 10 Ω.
A carbon resistor has bands red, red, brown, silver. Its value is
1. 2200 Ω ± 10%
2. 220 Ω ± 5%
3. 220 Ω ± 10%
4. 22 Ω ± 10%
Answer
C. 220 Ω ± 10%
Red 2, red 2, brown multiplier 10¹: 22 × 10 = 220 Ω. Silver means ± 10%.
As the temperature of a semiconductor rises, its resistivity
1. increases
2. decreases
3. remains constant
4. first increases, then becomes zero
Answer
B. decreases
More charge carriers are freed, so n rises and resistivity falls.
The SI unit of electron mobility is
1. m² V⁻¹ s⁻¹
2. m V⁻¹ s⁻¹
3. m s⁻¹
4. m² V s⁻¹
Answer
A. m² V⁻¹ s⁻¹
μ = v_d/E = (m s⁻¹)/(V m⁻¹) = m² V⁻¹ s⁻¹.
Kirchhoff's junction rule is based on the conservation of
1. momentum
2. mass
3. energy
4. charge
Answer
D. charge
The charge entering a junction equals the charge leaving it.
Given below are two statements: Statement I: The drift speed of electrons in a metal wire is of the order of 10⁻⁴ m s⁻¹. Statement II: A bulb lights almost at once after the switch is closed because the electric field is set up along the wire at nearly the speed of light. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both statements are correct.
Given below are two statements: Statement I: Resistivity depends on the material and the temperature. Statement II: Resistivity depends on the length of the wire. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Resistivity is a material property and does not depend on length or area.
Given below are two statements: Statement I: The V–I graph of a p–n junction diode is a straight line through the origin. Statement II: The V–I graph of a metallic conductor at constant temperature is a straight line through the origin. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
A diode is non-ohmic. A metal obeys Ohm's law at constant temperature.
Given below are two statements: Statement I: During discharge, the terminal voltage of a cell is greater than its emf. Statement II: During charging, the terminal voltage of a cell is less than its emf. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
V = ε – Ir in discharge (less than ε) and V = ε + Ir in charging (greater than ε). Both statements are wrong.
Which of the following statements are correct? 1. n identical cells in series have total emf nε. 2. n identical cells in parallel have total emf nε. 3. n identical cells in parallel have total internal resistance r/n.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Series: emf nε, resistance nr. Parallel: emf ε, resistance r/n. Statement 2 is false.
Given below are two statements: Statement I: The equivalent resistance of resistors in parallel is greater than the largest of the resistances. Statement II: In a series combination the same current flows through each resistor. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Parallel equivalent is less than the smallest resistance. Series current is the same.
Which of the following statements are correct? 1. The loop rule follows from the conservation of momentum. 2. The junction rule follows from the conservation of charge. 3. The loop rule follows from the conservation of energy.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
The loop rule is energy conservation; the junction rule is charge conservation.
Given below are two statements: Statement I: In a balanced Wheatstone bridge no current flows through the galvanometer. Statement II: The balance condition P/Q = R/S changes if the cell and galvanometer are interchanged. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
The balance condition stays the same when the cell and galvanometer positions are interchanged.
Given below are two statements: Statement I: Manganin has a large temperature coefficient of resistance. Statement II: Manganin is used for making standard resistors. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Manganin has a very small temperature coefficient, which is why it suits standard resistors.
Which of the following statements about the colour code of resistors are correct? 1. A gold tolerance band means ± 10%. 2. A silver tolerance band means ± 10%. 3. No tolerance band means ± 20%.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
Gold is ± 5%, silver is ± 10% and no band is ± 20%.
Assertion (A): Electrons drift very slowly in a wire, yet a lamp lights almost instantly when the switch is closed. Reason (R): The electric field is established along the entire wire at nearly the speed of light, so all free electrons start drifting together.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The field sets all electrons in motion at once; the drift speed itself stays small.
Assertion (A): The terminal voltage of a cell equals its emf when it supplies a current. Reason (R): The internal resistance of the cell causes a potential drop Ir inside the cell.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
V = ε – Ir is less than ε when I is not zero, so A is false. R is true.
Assertion (A): The resistivity of a wire is the same for a long wire and a short wire of the same material at the same temperature. Reason (R): The resistance of a wire is directly proportional to its length.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true. Resistivity is a material property; the length dependence belongs to resistance and does not explain it.
Assertion (A): The resistance of a wire increases when it is stretched. Reason (R): The resistivity of the material increases on stretching.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Length rises and area falls, so R rises. Resistivity is unchanged, so R is false.
Assertion (A): In a series combination, the current through each resistor is different. Reason (R): The potential drop across each resistor is proportional to its resistance.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Series resistors carry the same current, so A is false. R is true (V = IR with the same I).
Assertion (A): Nichrome is used for heating elements. Reason (R): Nichrome has a high resistivity and a high melting point.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
High resistivity gives large heating; the high melting point lets it stay red hot.
Match List I with List II. List I: a. Mobility b. Resistivity c. Conductivity d. Current density. List II: i. Ω m ii. S m⁻¹ iii. A m⁻² iv. m² V⁻¹ s⁻¹.
1. a-iv, b-i, c-ii, d-iii
2. a-i, b-iv, c-ii, d-iii
3. a-iv, b-i, c-iii, d-ii
4. a-ii, b-i, c-iv, d-iii
Answer
A. a-iv, b-i, c-ii, d-iii
Mobility m² V⁻¹ s⁻¹; resistivity Ω m; conductivity S m⁻¹; J in A m⁻².
Match List I with List II. List I: a. Series resistors b. Parallel resistors c. Power dissipated d. Terminal voltage of a cell. List II: i. ε – Ir ii. R₁ + R₂ iii. 1/R₁ + 1/R₂ iv. I²R.
1. a-iv, b-iii, c-ii, d-i
2. a-ii, b-iii, c-iv, d-i
3. a-iii, b-ii, c-iv, d-i
4. a-ii, b-iii, c-i, d-iv
Answer
B. a-ii, b-iii, c-iv, d-i
Series add; parallel add reciprocals; P = I²R; V = ε – Ir.
Match List I with List II. List I: a. Red b. Green c. Violet d. Grey. List II: i. 7 ii. 8 iii. 2 iv. 5.
1. a-iii, b-iv, c-i, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-iv, b-iii, c-i, d-ii
4. a-i, b-ii, c-iii, d-iv
Answer
A. a-iii, b-iv, c-i, d-ii
Colour code digits: red 2, green 5, violet 7, grey 8.
Match List I with List II. List I: a. Copper b. Silicon c. Manganin d. Nichrome. List II: i. Resistivity falls as temperature rises ii. Resistivity rises as temperature rises iii. Standard resistors, very small temperature coefficient iv. Heating elements.
1. a-iii, b-i, c-ii, d-iv
2. a-ii, b-i, c-iv, d-iii
3. a-ii, b-i, c-iii, d-iv
4. a-i, b-ii, c-iii, d-iv
Answer
C. a-ii, b-i, c-iii, d-iv
Metal: rises. Semiconductor: falls. Manganin: standard resistors. Nichrome: heaters.
Match List I with List II. List I: a. Emf of 4 cells of 1.5 V in series b. Internal resistance of 4 cells of 0.2 Ω in parallel c. 1 kWh d. Resistance of a 100 W, 200 V bulb. List II: i. 3.6 × 10⁶ J ii. 6 V iii. 400 Ω iv. 0.05 Ω.
1. a-ii, b-iv, c-iii, d-i
2. a-ii, b-iv, c-i, d-iii
3. a-i, b-iv, c-ii, d-iii
4. a-iv, b-ii, c-i, d-iii
Answer
B. a-ii, b-iv, c-i, d-iii
a: 4 × 1.5 = 6 V. b: 0.2/4 = 0.05 Ω. c: 1000 × 3600 = 3.6 × 10⁶ J. d: 200²/100 = 400 Ω.
Match List I with List II. List I: a. Junction rule b. Loop rule c. Balanced Wheatstone bridge d. Metre bridge. List II: i. P/Q = R/S ii. Conservation of energy iii. X = R(100 – l)/l iv. Conservation of charge.
1. a-i, b-ii, c-iv, d-iii
2. a-ii, b-iv, c-i, d-iii
3. a-iv, b-ii, c-iii, d-i
4. a-iv, b-ii, c-i, d-iii
Answer
D. a-iv, b-ii, c-i, d-iii
Junction: charge. Loop: energy. Balance: P/Q = R/S. Metre bridge: X = R(100 – l)/l.
A wire of cross-section 2 × 10⁻⁶ m² carries 3.2 A. The free electron density is 10²⁹ m⁻³ and e = 1.6 × 10⁻¹⁹ C. The drift speed is
1. 2 × 10⁻⁴ m s⁻¹
2. 1 × 10⁻³ m s⁻¹
3. 1 × 10⁻⁴ m s⁻¹
4. 1 × 10⁻⁵ m s⁻¹
Answer
C. 1 × 10⁻⁴ m s⁻¹
v = I/(neA) = 3.2/(10²⁹ × 1.6 × 10⁻¹⁹ × 2 × 10⁻⁶) = 3.2/(3.2 × 10⁴) = 10⁻⁴ m s⁻¹.
A bulb marked 100 W, 200 V is connected to a 100 V supply (resistance taken as constant). The power consumed is
1. 25 W
2. 100 W
3. 12.5 W
4. 50 W
Answer
A. 25 W
R = 200²/100 = 400 Ω. P = 100²/400 = 25 W. (Power falls with the square of voltage.)
Four identical cells, each of emf 1.5 V and internal resistance 0.5 Ω, are connected in series across a 4 Ω resistor. The current is
1. 3 A
2. 0.36 A
3. 1.5 A
4. 1 A
Answer
D. 1 A
ε = 6 V, r = 2 Ω. I = 6/(4 + 2) = 1 A.
A cell of 10 V and internal resistance 1 Ω is connected in series with a cell of 4 V and internal resistance 0.5 Ω, in opposition, and a 4.5 Ω resistor in one loop. The current is
1. 1.5 A
2. 2.33 A
3. 1 A
4. 0.6 A
Answer
C. 1 A
Net emf = 10 – 4 = 6 V. Total resistance = 1 + 0.5 + 4.5 = 6 Ω. I = 1 A.
In a bridge, arms AB = 2 Ω, BC = 4 Ω, AD = 3 Ω and DC = 6 Ω, with a 10 Ω galvanometer joined between B and D. A cell is joined across A and C. The effective resistance between A and C is
1. 3.6 Ω
2. 7.2 Ω
3. 15 Ω
4. 2.5 Ω
Answer
A. 3.6 Ω
2/4 = 3/6, so the bridge is balanced and the 10 Ω arm carries no current. R = (2 + 4)(3 + 6)/(6 + 9) = 54/15 = 3.6 Ω.
In a metre bridge, a 20 Ω resistor is in the left gap and the balance point is 25 cm from the left end. The resistance in the right gap is
1. 6.7 Ω
2. 60 Ω
3. 20 Ω
4. 80 Ω
Answer
B. 60 Ω
R/X = l/(100 – l) = 25/75, so X = 20 × 3 = 60 Ω.
A 10 Ω uniform wire is cut into five equal pieces, which are then joined in parallel. The equivalent resistance is
1. 2 Ω
2. 10 Ω
3. 0.08 Ω
4. 0.4 Ω
Answer
D. 0.4 Ω
Each piece is 2 Ω. Five 2 Ω in parallel give 2/5 = 0.4 Ω.

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