Class 12 · Physics · Part 1

Chapter 3: Current Electricity

Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.

3.1Electric currentCurrent, current density and drift

Current

Electric current is the rate of flow of charge through a cross-section: I = Q / t. The SI unit is the ampere (A), 1 A = 1 C s⁻¹.
By convention the direction of the current is the direction in which positive charge would flow. In a metal the carriers are electrons, which move the other way.
Current is a scalar (it adds algebraically), although it has a direction along the wire.
In a steady current the same charge passes every cross-section each second, so the current is the same at every point of a wire even if the cross-section changes.

Current density

The current density is the current per unit area perpendicular to the flow: j = I / A, in A m⁻². It is a vector, in the direction of the flow of positive charge.
In terms of the carriers, j = n q v_d, where n is the number density of carriers.

3.2Ohm's law and resistanceV = IR and resistivity

Ohm's law

Ohm's law: the current through a conductor is proportional to the potential difference across it, at constant temperature: V = I R.
The constant R is the resistance. Its unit is the ohm (Ω), 1 Ω = 1 V A⁻¹. The graph of V against I is a straight line through the origin.

V–I characteristics. A straight line through the origin means Ohm's law holds; a curved or one-sided graph means the device is non-ohmic.

Resistivity

For a conductor of length l and cross-section A: R = ρ l / A, where ρ is the resistivity of the material. Unit: Ω m.
The resistivity depends on the material and temperature, not on the shape. The conductivity is σ = 1 / ρ (unit S m⁻¹).
If a wire is stretched to n times its length at constant volume, the resistance becomes n² times.
The vector form of Ohm's law: j = σ E.

3.3Drift of electronsThe origin of resistivity

Free electrons and drift velocity

In a metal the free electrons move randomly with high speed, colliding with the ions. With no field the average velocity is zero.
In a field E each electron is accelerated by a = eE / m between collisions. The average time between collisions is the relaxation time τ.
The drift velocity is v_d = e E τ / m, opposite to E. It is small, about 10⁻⁴ m s⁻¹ in a copper wire, although the electric signal travels almost at the speed of light.
Current: I = n e A v_d, where n is the number of free electrons per unit volume.
Mobility μ = v_d / E (unit m² V⁻¹ s⁻¹).
Comparing with Ohm's law: ρ = m / (n e² τ).

Drift of electrons. Between collisions an electron is accelerated opposite to E; collisions randomise the motion, leaving a small average drift velocity.

3.4Limitations of Ohm's lawOhmic and non-ohmic devices

Where Ohm's law fails

A device is ohmic if V is proportional to I. Metals at constant temperature are ohmic.
In a non-ohmic device the V–I graph is not a straight line, or it depends on the direction of the current. Examples are a diode, a thermistor and a semiconductor junction.
In a diode the current is large in one direction and almost zero in the other.
In some materials the relation between V and I depends on the sign of V, or I is not a single-valued function of V.

3.5Resistivity of materialsConductors, semiconductors and insulators

Typical resistivities

Metals (copper, silver): very low resistivity, about 10⁻⁸ to 10⁻⁶ Ω m.
Insulators (glass, rubber): very high resistivity, 10¹⁰ Ω m and above.
Semiconductors (silicon, germanium): in between, and the resistivity falls steeply as temperature rises or with impurities.
Alloys such as nichrome have a higher resistivity than the metals they contain, which is why they are used in heating elements. Manganin and constantan have very small temperature coefficients and are used in standard resistors.

Temperature dependence

For a metallic conductor over a modest range: ρ_T = ρ₀ [1 + α (T − T₀)], and similarly R_T = R₀ [1 + α (T − T₀)].
The temperature coefficient of resistivity α is positive for metals (resistance rises with temperature, since collisions become more frequent), very small for manganin, and negative for semiconductors and insulators.
Resistivity of a metal rises with temperature because the relaxation time τ decreases.

3.6Combination of resistorsSeries and parallel

Series

The same current flows through each resistor; the voltages add. R = R₁ + R₂ + R₃. The total is larger than the largest resistor.
The voltage divides in proportion to the resistances: V₁ = V R₁ / (R₁ + R₂).

Parallel

The voltage across each is the same; the currents add. 1/R = 1/R₁ + 1/R₂ + 1/R₃. The total is smaller than the smallest resistor.
Two resistors: R = R₁R₂ / (R₁ + R₂). The current divides inversely as the resistance: I₁ = I R₂ / (R₁ + R₂).
For n equal resistors R: series nR, parallel R/n.

Resistors in series carry the same current; resistors in parallel have the same voltage.

3.7Electrical energy and powerJoule heating

Power and heat

When a current flows through a resistor, the energy supplied by the source is turned into heat. The power is P = V I = I² R = V² / R, in watts (W).
Energy in time t: E = P t (joules). The commercial unit is the kilowatt-hour (kWh): 1 kWh = 3.6 × 10⁶ J.
A bulb or heater rated for V₀ and P₀ has resistance R = V₀² / P₀. At a different voltage V it uses P = P₀ (V/V₀)².
In transmission, P = VI fixed, so a high voltage means a small current and a small I²R loss in the cables.

3.8Cells, emf and internal resistanceTerminal voltage

emf and terminal voltage

The emf ε of a cell is the potential difference between its terminals when no current is drawn (an open circuit). It is the work done per unit charge by the cell.
The electrolyte offers an internal resistance r. When a current I is drawn the terminal voltage is V = ε − I r (less than ε).
With an external resistance R: I = ε / (R + r) and V = I R.
The maximum current (short circuit, R = 0) is ε / r.
When a cell is being charged by a source, the current flows into its positive terminal and V = ε + I r (the terminal voltage is larger than the emf). A series resistor limits the charging current.
Maximum power is delivered to the load when R = r, and it is ε² / 4r.

A cell drives current through the external resistance R and its own internal resistance r. The terminal voltage is V = ε − Ir.

3.9Cells in series and parallelCombining sources

Series

Cells connected positive to negative: ε_eq = ε₁ + ε₂ and r_eq = r₁ + r₂.
If one cell is connected the other way round (opposing): ε_eq = ε₁ − ε₂ (with r_eq = r₁ + r₂).
For n identical cells in series: nε and nr.

Parallel

For two cells in parallel (like terminals together): ε_eq = (ε₁r₂ + ε₂r₁) / (r₁ + r₂) and r_eq = r₁r₂ / (r₁ + r₂).
For n identical cells in parallel: the emf is ε and the internal resistance is r/n.
Series suits a large external resistance; parallel suits a small one.

3.10Kirchhoff's rulesAnalysing networks

The two rules

Junction rule: at any junction the sum of the currents entering equals the sum leaving. This expresses conservation of charge.
Loop rule: the algebraic sum of the changes in potential round any closed loop is zero. This expresses conservation of energy.

Kirchhoff's rules: the currents at a junction add up to zero, and the potential changes round a closed loop add up to zero.

Sign conventions for the loop rule

Going through a resistor in the direction of the current: the change is −IR; against the current, +IR.
Going through a cell from its negative to its positive terminal: +ε; from positive to negative: −ε (ignoring internal resistance).
Choose a direction of current in each branch. If a calculated current is negative, it actually flows the opposite way.
For a network with n unknown currents write one equation per junction (using n−1 of them) and the remaining ones from loops.

3.11Wheatstone bridgeComparing resistances

The bridge

Four resistors P, Q, R, S form a diamond with a galvanometer between the two middle junctions. When the galvanometer shows no deflection the bridge is balanced: P / Q = R / S.
If three resistances are known, the fourth is found: S = Q R / P. The result does not depend on the emf or on the galvanometer.
At balance the two middle junctions are at the same potential, so the arm containing the galvanometer can be removed without changing the rest of the circuit.

Wheatstone bridge. With no current through G the bridge is balanced: P/Q = R/S.

Meter bridge

In a meter bridge a 1 m uniform wire is used as two of the arms. If the balance point is at l cm from one end, then R / S = l / (100 − l), so S = R (100 − l) / l.
The wire is uniform, so its resistance is proportional to its length.

Quick referenceFormulas at a glanceRevise before the exam

Key results

Current: I = Q/t = n e A v_d; j = I/A
Ohm's law: V = IR; R = ρl/A; ρ = m/(ne²τ)
Temperature: R_T = R₀[1 + α(T − T₀)]
Series: R = ΣRᵢ; parallel: 1/R = Σ1/Rᵢ
Power: P = VI = I²R = V²/R; 1 kWh = 3.6 × 10⁶ J
Cell: V = ε − Ir; I = ε/(R + r); max power at R = r
Kirchhoff: ΣI = 0 at a junction; ΣΔV = 0 round a loop
Wheatstone: P/Q = R/S; meter bridge S = R(100 − l)/l

Common mistakes

Forgetting the internal resistance when a current is drawn from the cell.
Using ε − Ir while a cell is being charged (it should be ε + Ir).
Wrong sign for a resistor or a cell in the loop equation.
Mixing mm² or cm² with m² in R = ρl/A.

Try each problem first, then open the solution to check your working.

3.1Current and driftCharge, current density and drift speed · 6 problems

Q1Charge and electrons

A current of 2 A flows for 5 minutes. Find the charge that passes and the number of electrons.

Show solutionHide solution

Given: I = 2 A; t = 300 s

Use: Q = I t and n = Q / e

Q = 2 × 300 = 600 C

n = 600 / 1.6 × 10⁻¹⁹ = 3.75 × 10²¹

Answer: 600 C; 3.75 × 10²¹ electrons

Q2Current from electrons

5 × 10¹⁸ electrons pass through a cross-section of a wire every second. Find the current.

Show solutionHide solution

Given: n = 5 × 10¹⁸ per second

Use: I = n e / t

I = 5 × 10¹⁸ × 1.6 × 10⁻¹⁹ = 0.8 A

Answer: 0.8 A

Q3Drift speed

A copper wire of cross-section 1.0 mm² carries a current of 2 A. Find the drift speed of the electrons. (n = 8.5 × 10²⁸ m⁻³)

Show solutionHide solution

Given: I = 2 A; A = 10⁻⁶ m²; n = 8.5 × 10²⁸ m⁻³

Use: v_d = I / (n e A)

v_d = 2 / (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 10⁻⁶) = 2 / 13600 = 1.47 × 10⁻⁴ m/s

Answer: 1.5 × 10⁻⁴ m/s (about 0.15 mm/s)

Q4Current density

A wire of diameter 2 mm carries 5 A. Find the current density.

Show solutionHide solution

Given: I = 5 A; r = 10⁻³ m

Use: j = I / A

A = π r² = 3.14 × 10⁻⁶ m²

j = 5 / 3.14 × 10⁻⁶ = 1.59 × 10⁶ A/m²

Answer: 1.6 × 10⁶ A/m²

Q5Mobility

In a field of 2 V/m the drift speed of electrons in a conductor is 4 × 10⁻³ m/s. Find their mobility.

Show solutionHide solution

Given: E = 2 V/m; v_d = 4 × 10⁻³ m/s

Use: μ = v_d / E

μ = 4 × 10⁻³ / 2 = 2 × 10⁻³ m² V⁻¹ s⁻¹

Answer: 2 × 10⁻³ m² V⁻¹ s⁻¹

Q6Relaxation time

The resistivity of copper is 1.7 × 10⁻⁸ Ω m and n = 8.5 × 10²⁸ m⁻³. Estimate the relaxation time of its electrons.

Show solutionHide solution

Given: ρ = 1.7 × 10⁻⁸ Ω m; n = 8.5 × 10²⁸ m⁻³

Use: τ = m / (n e² ρ)

n e² ρ = 8.5 × 10²⁸ × (1.6 × 10⁻¹⁹)² × 1.7 × 10⁻⁸ = 3.70 × 10⁻¹⁷

τ = 9.11 × 10⁻³¹ / 3.70 × 10⁻¹⁷ = 2.46 × 10⁻¹⁴ s

Answer: τ ≈ 2.5 × 10⁻¹⁴ s

3.2Resistance and resistivityWires, temperature and measurement · 6 problems

Q7Resistance of a wire

Find the resistance of a copper wire 10 m long and 0.5 mm² in cross-section. (ρ = 1.7 × 10⁻⁸ Ω m)

Show solutionHide solution

Given: l = 10 m; A = 5 × 10⁻⁷ m²

Use: R = ρ l / A

R = 1.7 × 10⁻⁸ × 10 / 5 × 10⁻⁷ = 0.34 Ω

Answer: 0.34 Ω

Q8Stretching a wire

A wire of resistance 4 Ω is stretched to twice its length. Find the new resistance. Assume the volume is unchanged.

Show solutionHide solution

Given: R = 4 Ω; l′ = 2l

Use: R = ρ l / A with l A = constant

Length × 2 means area ÷ 2.

R′ = ρ (2l) / (A/2) = 4 ρ l / A = 4R = 16 Ω

Answer: 16 Ω

Q9Cutting and joining a wire

A wire of resistance 12 Ω is cut into three equal pieces, which are then joined in parallel. Find the resistance of the combination.

Show solutionHide solution

Given: R = 12 Ω

Use: Each piece is a third of the length, so a third of the resistance

Each piece: 12 / 3 = 4 Ω

In parallel: 1/R = 1/4 + 1/4 + 1/4 = 3/4, so R = 4/3 = 1.33 Ω

Answer: 1.33 Ω (4/3 Ω)

Q10Resistance at a higher temperature

A coil has a resistance of 50 Ω at 20 °C. Find its resistance at 120 °C. The temperature coefficient is 0.004 °C⁻¹.

Show solutionHide solution

Given: R₀ = 50 Ω at 20 °C; α = 0.004 °C⁻¹

Use: R = R₀ [1 + α (T − T₀)]

R = 50 [1 + 0.004 × 100] = 50 × 1.4 = 70 Ω

Answer: 70 Ω

Q11Finding α

The resistance of a wire is 40 Ω at 20 °C and 52 Ω at 100 °C. Find the temperature coefficient of resistance.

Show solutionHide solution

Given: R₁ = 40 Ω at 20 °C; R₂ = 52 Ω at 100 °C

Use: α = (R₂ − R₁) / [R₁ (T₂ − T₁)]

α = (52 − 40) / (40 × 80) = 12 / 3200 = 3.75 × 10⁻³ °C⁻¹

Answer: 3.75 × 10⁻³ °C⁻¹

Q12Resistivity from measurements

A 2 m wire of diameter 0.4 mm carries 0.3 A when the potential difference across it is 1.5 V. Find the resistivity.

Show solutionHide solution

Given: l = 2 m; d = 0.4 mm; V = 1.5 V; I = 0.3 A

Use: R = V/I and ρ = R A / l

R = 1.5 / 0.3 = 5 Ω

A = π (0.2 × 10⁻³)² = 1.26 × 10⁻⁷ m²

ρ = 5 × 1.26 × 10⁻⁷ / 2 = 3.14 × 10⁻⁷ Ω m

Answer: 3.1 × 10⁻⁷ Ω m

3.3Resistor combinations and bridgesSeries, parallel, Wheatstone and meter bridge · 6 problems

Q13A series–parallel combination

A 3 Ω and a 6 Ω resistor are in parallel and this pair is in series with a 4 Ω resistor. Find the total resistance, and the current if the combination is connected to 12 V.

Show solutionHide solution

Given: R₁ = 3 Ω; R₂ = 6 Ω; R₃ = 4 Ω; V = 12 V

Use: Combine the parallel pair first

Parallel pair: (3 × 6) / (3 + 6) = 2 Ω

Total: 2 + 4 = 6 Ω

I = V / R = 12 / 6 = 2 A

Answer: 6 Ω; 2 A

Q14Potential difference across the parallel pair

In the previous circuit find the potential difference across the parallel pair and the current in the 3 Ω resistor.

Show solutionHide solution

Given: I = 2 A; parallel resistance 2 Ω

Use: Voltage across the parallel pair = I × R_parallel

V = 2 × 2 = 4 V

Current in 3 Ω: 4 / 3 = 1.33 A (and 4 / 6 = 0.67 A in 6 Ω; total 2 A ✓)

Answer: 4 V; 1.33 A

Q15Four equal resistors

Four resistors of 10 Ω each are connected (a) in series and (b) in parallel. Find the ratio of the two equivalent resistances.

Show solutionHide solution

Given: R = 10 Ω; n = 4

Use: Series: nR; parallel: R/n

Series: 40 Ω; parallel: 10 / 4 = 2.5 Ω

Ratio = 40 / 2.5 = 16 (= n²)

Answer: 40 Ω and 2.5 Ω; ratio 16

Q16Wheatstone bridge

In a Wheatstone bridge P = 10 Ω, Q = 15 Ω and R = 20 Ω. Find S for balance.

Show solutionHide solution

Given: P = 10; Q = 15; R = 20 Ω

Use: P/Q = R/S

S = R Q / P = 20 × 15 / 10 = 30 Ω

Answer: 30 Ω

Q17Meter bridge

In a meter bridge the known resistance is 10 Ω and the balance point is 40 cm from the end where it is connected. Find the unknown resistance.

Show solutionHide solution

Given: R = 10 Ω; l = 40 cm

Use: S = R (100 − l) / l

S = 10 × 60 / 40 = 15 Ω

Answer: 15 Ω

Q18A balanced bridge

A bridge has arms of 2 Ω, 4 Ω, 3 Ω and 6 Ω so that 2/4 = 3/6, and a 100 Ω galvanometer across the middle. Find the equivalent resistance between the input corners.

Show solutionHide solution

Given: P = 2, Q = 4, R = 3, S = 6 Ω

Use: At balance no current flows through the galvanometer, so ignore it

Upper path: 2 + 4 = 6 Ω; lower path: 3 + 6 = 9 Ω

In parallel: (6 × 9) / 15 = 3.6 Ω

Answer: 3.6 Ω

3.4Cells and Kirchhoff's rulesemf, internal resistance, combinations and networks · 7 problems

Q19Terminal voltage

A cell of emf 12 V and internal resistance 0.5 Ω is connected to a 5.5 Ω resistor. Find the current and the terminal voltage.

Show solutionHide solution

Given: ε = 12 V; r = 0.5 Ω; R = 5.5 Ω

Use: I = ε/(R + r); V = ε − Ir

I = 12 / 6 = 2 A

V = 12 − 2 × 0.5 = 11 V (= I R = 2 × 5.5 ✓)

Answer: 2 A; 11 V

Q20Short-circuit current

A cell of emf 6 V and internal resistance 0.2 Ω is short-circuited. Find the current.

Show solutionHide solution

Given: ε = 6 V; r = 0.2 Ω; R = 0

Use: I = ε / r

I = 6 / 0.2 = 30 A

Answer: 30 A

Q21Cells in series (aiding)

Cells of emf 2 V (r = 0.5 Ω) and 1.5 V (r = 0.3 Ω) are connected in series, helping each other, to a 3.2 Ω resistor. Find the current.

Show solutionHide solution

Given: ε₁ = 2 V, r₁ = 0.5 Ω; ε₂ = 1.5 V, r₂ = 0.3 Ω; R = 3.2 Ω

Use: ε = ε₁ + ε₂; r = r₁ + r₂

ε = 3.5 V; r = 0.8 Ω

I = 3.5 / (3.2 + 0.8) = 0.875 A

Answer: 0.875 A

Q22Cells in series (opposing)

The same two cells are connected in series with opposite polarity to the 3.2 Ω resistor. Find the current.

Show solutionHide solution

Given: Same cells; ε = 2 − 1.5

Use: ε = ε₁ − ε₂

ε = 0.5 V; r = 0.8 Ω

I = 0.5 / 4.0 = 0.125 A

Answer: 0.125 A, in the direction set by the larger cell

Q23Cells in parallel

Four identical cells, each of emf 1.5 V and internal resistance 0.5 Ω, are connected in parallel to a 2.875 Ω resistor. Find the current.

Show solutionHide solution

Given: n = 4; ε = 1.5 V; r = 0.5 Ω

Use: Parallel: emf stays ε, resistance r/n

Internal resistance = 0.5 / 4 = 0.125 Ω

I = 1.5 / (2.875 + 0.125) = 0.5 A

Answer: 0.5 A

Q24Two batteries and three resistors

A 12 V battery with a 4 Ω resistor and a 6 V battery with a 2 Ω resistor are connected, each in its own branch, to a common 3 Ω resistor (both batteries send current into it). Find the current in each branch.

Show solutionHide solution

Given: ε₁ = 12 V with 4 Ω; ε₂ = 6 V with 2 Ω; middle resistor 3 Ω

Use: Junction rule: I₃ = I₁ + I₂. Loop rule on each loop

Loop 1: 12 − 4 I₁ − 3 (I₁ + I₂) = 0, so 7 I₁ + 3 I₂ = 12

Loop 2: 6 − 2 I₂ − 3 (I₁ + I₂) = 0, so 3 I₁ + 5 I₂ = 6

Solving: I₁ = 21/13 = 1.62 A, I₂ = 3/13 = 0.23 A

I₃ = I₁ + I₂ = 24/13 = 1.85 A

Answer: 1.62 A, 0.23 A and 1.85 A in the middle resistor

Q25Charging a cell

A cell of emf 6 V and internal resistance 0.5 Ω is charged from a 12 V supply through a 5 Ω resistor. Find the charging current and the terminal voltage of the cell.

Show solutionHide solution

Given: ε = 6 V; r = 0.5 Ω; supply 12 V; R = 5 Ω

Use: I = (net emf) / (total resistance); V = ε + Ir

I = (12 − 6) / (5 + 0.5) = 6 / 5.5 = 1.09 A

V = 6 + 1.09 × 0.5 = 6.55 V (more than the emf)

Answer: 1.09 A; 6.55 V

3.5Power and energyJoule heating and electrical energy · 6 problems

Q26Heater

A heater is rated 1 kW at 230 V. Find its resistance and the current it draws.

Show solutionHide solution

Given: P = 1000 W; V = 230 V

Use: R = V²/P; I = P/V

R = 230² / 1000 = 52.9 Ω

I = 1000 / 230 = 4.35 A

Answer: 52.9 Ω; 4.35 A

Q27Cost of electricity

A 2 kW heater runs for 3 hours every day for 30 days. Find the energy used and the cost at ₹6 per unit.

Show solutionHide solution

Given: P = 2 kW; 3 h/day; 30 days

Use: E = P t

Energy = 2 × 3 × 30 = 180 kWh (units)

Cost = 180 × 6 = ₹1080

Answer: 180 kWh; ₹1080

Q28A bulb on a lower voltage

A bulb rated 100 W, 220 V is connected to a 110 V supply. Find the power it uses (ignore change in resistance).

Show solutionHide solution

Given: P₀ = 100 W; V₀ = 220 V; V = 110 V

Use: P = P₀ (V / V₀)²

P = 100 × (110/220)² = 100 × ¼ = 25 W

Answer: 25 W

Q29Heating water

A 1 kW immersion heater heats 2 kg of water from 20 °C to 70 °C. How long does it take? (Specific heat of water 4200 J kg⁻¹ K⁻¹; ignore losses.)

Show solutionHide solution

Given: P = 1000 W; m = 2 kg; ΔT = 50 K

Use: P t = m c ΔT

Heat needed = 2 × 4200 × 50 = 4.2 × 10⁵ J

t = 4.2 × 10⁵ / 1000 = 420 s = 7 minutes

Answer: 7 minutes

Q30Maximum power

A battery of emf 12 V and internal resistance 2 Ω supplies a variable external resistor. For what resistance is the power delivered maximum, and what is that power?

Show solutionHide solution

Given: ε = 12 V; r = 2 Ω

Use: Maximum power when R = r: P_max = ε² / 4r

R = 2 Ω

P_max = 144 / 8 = 18 W

Answer: R = 2 Ω; P_max = 18 W

Q31Bulbs in series

Two bulbs, 60 W and 100 W, both rated for 220 V, are joined in series across a 220 V supply. Which glows brighter?

Show solutionHide solution

Given: R = V²/P for each

Use: In series the same current flows, so the larger R gets more power: P = I²R

R₁ (60 W) = 220² / 60 = 807 Ω; R₂ (100 W) = 484 Ω

The same current flows, so power ∝ R. The 60 W bulb has the higher resistance.

Answer: The 60 W bulb glows brighter when they are in series

Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.

3.1Exercises 3.1 to 3.9Cells, resistivity, networks and drift · 9 questions

Q13.1 The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Show answerHide answer

The current is maximum when the external resistance is zero (the terminals are shorted): I_max = ε / r = 12 / 0.4 = 30 A.

Q23.2 A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Show answerHide answer

I = ε / (R + r), so R + r = 10 / 0.5 = 20 Ω and R = 20 − 3 = 17 Ω.

Terminal voltage V = ε − Ir = 10 − 0.5 × 3 = 8.5 V (check: I R = 0.5 × 17 = 8.5 V).

Q33.3 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10⁻⁴ °C⁻¹?

Show answerHide answer

R_T = R₀ [1 + α (T − T₀)]. 117 = 100 [1 + 1.7 × 10⁻⁴ (T − 27)]

1.17 − 1 = 1.7 × 10⁻⁴ (T − 27), so T − 27 = 0.17 / 1.7 × 10⁻⁴ = 1000.

T = 1027 °C.

Q43.4 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m², and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Show answerHide answer

ρ = R A / l = (5.0 × 6.0 × 10⁻⁷) / 15 = 2.0 × 10⁻⁷ Ω m.

Q53.5 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Show answerHide answer

α = (R₂ − R₁) / [R₁ (T₂ − T₁)] = (2.7 − 2.1) / [2.1 × (100 − 27.5)] = 0.6 / 152.25

α = 3.9 × 10⁻³ °C⁻¹.

Q63.6 A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10⁻⁴ °C⁻¹.

Show answerHide answer

Initial resistance (at 27 °C): R₁ = 230 / 3.2 = 71.9 Ω. Steady resistance: R₂ = 230 / 2.8 = 82.1 Ω.

R₂ = R₁ [1 + α (T₂ − 27)], so T₂ − 27 = (R₂/R₁ − 1) / α = (1.1429 − 1) / 1.7 × 10⁻⁴ = 840 °C.

T₂ = 27 + 840 ≈ 867 °C.

Q73.7 Determine the current in each branch of the network shown in Fig. 3.20. (The network: AB = 10 Ω, BC = 5 Ω, AD = 5 Ω, DC = 10 Ω, BD = 5 Ω, with a 10 V cell and a 10 Ω resistor in series between C and A.)

Show answerHide answer

Let the current in AB be I₁, in AD be I₂ (total I₁ + I₂ leaves the cell), and in BD be I₃ (from B to D). Then BC carries I₁ − I₃ and DC carries I₂ + I₃.

Loop ABDA: 10 I₁ + 5 I₃ − 5 I₂ = 0 → 2 I₁ + I₃ − I₂ = 0

Loop BCDB: 5 (I₁ − I₃) − 10 (I₂ + I₃) − 5 I₃ = 0 → I₁ − 2 I₂ − 4 I₃ = 0

Loop through the cell (A → B → C → cell → A): 10 I₁ + 5 (I₁ − I₃) + 10 (I₁ + I₂) = 10 → 5 I₁ + 2 I₂ − I₃ = 2 (after dividing by 5)

Solving the three equations: I₁ = 4/17 A, I₂ = 6/17 A, I₃ = −2/17 A.

Current in each branch: AB = 4/17 A, AD = 6/17 A, BC = 6/17 A, DC = 4/17 A, and in BD = 2/17 A flowing from D to B (the negative sign shows that it flows opposite to the assumed direction). The total current from the cell is 10/17 A.

Check: the bridge between A and C has resistance 7 Ω, which with the 10 Ω in series makes 17 Ω, so I = 10/17 A ✓.

Q83.8 A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Show answerHide answer

Effective emf in the circuit = 120 − 8 = 112 V; total resistance = 15.5 + 0.5 = 16 Ω. Charging current I = 112 / 16 = 7 A.

During charging the terminal voltage of the battery is V = ε + Ir = 8 + 7 × 0.5 = 11.5 V.

Purpose of the series resistor: without it the current would be (120 − 8)/0.5 = 224 A, which would damage the battery. The resistor limits the charging current to a safe value.

Q93.9 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ m⁻³. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10⁻⁶ m² and it is carrying a current of 3.0 A.

Show answerHide answer

Drift speed v_d = I / (n e A) = 3.0 / (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.0 × 10⁻⁶) = 3.0 / 27200 = 1.1 × 10⁻⁴ m/s.

Time = length / v_d = 3.0 / 1.1 × 10⁻⁴ = 2.7 × 10⁴ s, about 7.6 hours.

The electrons drift very slowly. The bulb lights at once because the electric field is set up almost instantly (at nearly the speed of light) and moves all the electrons in the wire together.