Exercises
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?
Answer
Maximum current flows when the external resistance is zero (terminals shorted).
I_max = ε/r → 12/0.4 → I_max = 30 A
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?
Answer
I = ε/(R + r)
⇒ R + r = 10/0.5 = 20 Ω
⇒ R = 20 – 3 = 17 Ω
Terminal voltage V = ε – Ir
⇒ 10 – 0.5 × 3
⇒ V = 8.5 V (equivalently V = IR = 0.5 × 17 = 8.5 V)
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.
Answer
R_T = R₀[1 + α(T – T₀)]
⇒ 117 = 100[1 + 1.70 × 10⁻⁴ (T – 27)]
T – 27 = 0.17/(1.70 × 10⁻⁴)
⇒ T – 27 = 1000
⇒ T = 1027 °C
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?
Answer
R = ρl/A
⇒ ρ = RA/l
⇒ ρ = (5.0 × 6.0 × 10⁻⁷)/15
⇒ ρ = 2.0 × 10⁻⁷ Ω m
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.
Answer
α = (R₂ – R₁)/[R₁(T₂ – T₁)]
⇒ α = (2.7 – 2.1)/[2.1 × (100 – 27.5)]
⇒ α = 0.6/152.25
⇒ α ≈ 3.9 × 10⁻³ °C⁻¹
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.
Answer
At room temperature: R₁ = 230/3.2 = 71.9 Ω. At steady temperature: R₂ = 230/2.8 = 82.1 Ω.
T₂ – T₁ = (R₂ – R₁)/(αR₁)
⇒ (82.1 – 71.9)/(1.70 × 10⁻⁴ × 71.9)
⇒ 10.27/0.01222
⇒ ≈ 840 °C
T₂ = 840 + 27 ≈ 867 °C
7. Determine the current in each branch of the network shown in Fig. 3.20.
Answer
(Network: A–B = 10 Ω, B–C = 5 Ω, A–D = 5 Ω, D–C = 10 Ω, B–D = 5 Ω; a 10 V battery with a 10 Ω series resistor is connected between C and A, driving current into A.)
Ans. Let I = total current, I₁ = current in AB, I – I₁ in AD, I₂ = current in BD (B to D). Then BC carries I₁ – I₂ and DC carries I – I₁ + I₂.
Loop ABDA (Kirchhoff's loop rule): 10I₁ + 5I₂ – 5(I – I₁) = 0
⇒ 3I₁ + I₂ = I … (1)
Loop BCDB: 5(I₁ – I₂) – 10(I – I₁ + I₂) – 5I₂ = 0
⇒ 3I₁ – 4I₂ = 2I … (2)
(1) – (2): 5I₂ = –I
⇒ I₂ = –I/5. Then from (1): 3I₁ = 6I/5
⇒ I₁ = 2I/5.
Outer loop through the battery (A–B–C–battery–A): 10I₁ + 5(I₁ – I₂) + 10I = 10
⇒ 4I + 3I + 10I = 10
⇒ 17I = 10
⇒ I = 10/17 A
Results: AB = 4/17 A; AD = 6/17 A; BC = 6/17 A; DC = 4/17 A; BD = 2/17 A, flowing from D to B (the negative sign of I₂ shows the assumed direction B
⇒ D is reversed); main branch (battery and 10 Ω) = 10/17 A.
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?
Answer
During charging the battery's emf opposes the supply. I = (120 – 8)/(15.5 + 0.5)
⇒ 112/16
⇒ I = 7 A
Terminal voltage while charging V = ε + Ir
⇒ 8 + 7 × 0.5
⇒ V = 11.5 V
The series resistor limits the charging current. Without it, the current would be (120 – 8)/0.5 = 224 A, which would overheat and damage the battery and the supply.
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.
Answer
I = neAv_d
⇒ v_d = I/(neA)
⇒ v_d = 3.0/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.0 × 10⁻⁶)
⇒ v_d = 3.0/(2.72 × 10⁴)
⇒ v_d ≈ 1.1 × 10⁻⁴ m/s
t = l/v_d
⇒ 3.0/(1.1 × 10⁻⁴)
⇒ t ≈ 2.7 × 10⁴ s ≈ 7.5 hours
Important Extra Questions
Define drift velocity and write its relation with current.
Answer
Drift velocity is the average velocity acquired by free electrons in a conductor under an applied field: v_d = –eEτ/m. Current I = neAv_d, where τ is the average relaxation time.
Define mobility. Give its SI unit.
Answer
Mobility μ is the magnitude of drift velocity per unit electric field: μ = |v_d|/E = eτ/m. SI unit: m² V⁻¹ s⁻¹.
How does the resistivity of a metal and of a semiconductor change with temperature?
Answer
For metals, resistivity increases with temperature because collisions become more frequent (τ decreases). For semiconductors, resistivity decreases with temperature because the number density of charge carriers n increases sharply.
State Kirchhoff's rules.
Answer
Junction rule: the sum of currents entering a junction equals the sum leaving it (charge conservation). Loop rule: the algebraic sum of potential changes around any closed loop is zero (energy conservation).
Two cells of emf ε₁, ε₂ and internal resistances r₁, r₂ are connected in parallel. Write the equivalent emf and internal resistance.
Answer
ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) and r_eq = r₁r₂/(r₁ + r₂). For a series connection, ε_eq = ε₁ + ε₂ and r_eq = r₁ + r₂.