Q3.1 The storage battery of a car has an e.m.f. 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?
Given:
E = 12 V
r = 0.4 Ω
Maximum current is drawn when external resistance R = 0.
Using relation:
I = E / r
Substituting values:
I = 12 / 0.4
I = 30 A
Maximum current = 30 A
Q3.2 A battery of e.m.f. 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, find (a) the resistance of the resistor (b) the terminal voltage of the battery.
Given:
E = 10 V
r = 3 Ω
I = 0.5 A
(a) Using:
E = I(R + r)
10 = 0.5 (R + 3)
10 = 0.5R + 1.5
0.5R = 8.5
R = 17 Ω
(b) Terminal voltage:
V = E − Ir
V = 10 − (0.5 × 3)
V = 10 − 1.5
V = 8.5 V
Resistance = 17 Ω
Terminal voltage = 8.5 V
Q3.3 (a) Three resistors 1 Ω, 2 Ω and 3 Ω are connected in series. Find total resistance. (b) If connected to 12 V battery, find potential drop across each resistor.
(a) In series:
R_total = R₁ + R₂ + R₃
R_total = 1 + 2 + 3 = 6 Ω
(b) Current:
I = V / R_total = 12 / 6 = 2 A
Voltage drops:
V₁ = IR₁ = 2 × 1 = 2 V
V₂ = IR₂ = 2 × 2 = 4 V
V₃ = IR₃ = 2 × 3 = 6 V
Total resistance = 6 Ω
Voltage drops = 2 V, 4 V, 6 V
Q3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are connected in parallel. Find total resistance. (b) If connected to 20 V battery, find current through each resistor and total current.
(a) In parallel:
1/R = 1/2 + 1/4 + 1/5
1/R = 0.5 + 0.25 + 0.2 = 0.95
R = 1 / 0.95 ≈ 1.05 Ω
(b) Current through each resistor:
I₁ = V/R₁ = 20/2 = 10 A
I₂ = 20/4 = 5 A
I₃ = 20/5 = 4 A
Total current:
I_total = I₁ + I₂ + I₃ = 19 A
Equivalent resistance ≈ 1.05 Ω
Currents = 10 A, 5 A, 4 A
Total current = 19 A
Q3.5 At room temperature (27°C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance becomes 117 Ω? (α = 1.70 × 10⁻⁴ °C⁻¹)
Given:
R₀ = 100 Ω
R = 117 Ω
T₀ = 27°C
α = 1.70 × 10⁻⁴ °C⁻¹
Using formula:
R = R₀ [1 + α(T − T₀)]
Substitute:
117 = 100 [1 + 1.70×10⁻⁴ (T − 27)]
Divide by 100:
1.17 = 1 + 1.70×10⁻⁴ (T − 27)
0.17 = 1.70×10⁻⁴ (T − 27)
T − 27 = 0.17 / (1.70×10⁻⁴)
T − 27 = 1000
T = 1027°C
Temperature of element ≈ 1027°C
Q3.5 At room temperature (27°C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance becomes 117 Ω? (α = 1.70 × 10⁻⁴ °C⁻¹)
Given:
R₀ = 100 Ω
R = 117 Ω
T₀ = 27°C
α = 1.70 × 10⁻⁴ °C⁻¹
Using formula:
R = R₀ [1 + α(T − T₀)]
Substitute:
117 = 100 [1 + 1.70×10⁻⁴ (T − 27)]
Divide by 100:
1.17 = 1 + 1.70×10⁻⁴ (T − 27)
0.17 = 1.70×10⁻⁴ (T − 27)
T − 27 = 0.17 / (1.70×10⁻⁴)
T − 27 = 1000
T = 1027°C
Temperature of element ≈ 1027°C
Q3.7 A silver wire has resistance 2.1 Ω at 27.5°C and 2.7 Ω at 100°C. Find temperature coefficient of resistance.
Given:
R₁ = 2.1 Ω at T₁ = 27.5°C
R₂ = 2.7 Ω at T₂ = 100°C
Using:
R₂ = R₁ [1 + α (T₂ − T₁)]
Substitute:
2.7 = 2.1 [1 + α(72.5)]
2.7 / 2.1 = 1 + 72.5α
1.286 = 1 + 72.5α
0.286 = 72.5α
α = 0.286 / 72.5
α ≈ 3.95 × 10⁻³ °C⁻¹
Temperature coefficient ≈ 3.95 × 10⁻³ °C⁻¹
Q3.8 A heating element connected to 230 V supply draws 3.2 A initially and 2.8 A when hot. Room temperature is 27°C. (α = 1.70 × 10⁻⁴ °C⁻¹) Find steady temperature.
Initial resistance:
R₀ = V/I₀ = 230 / 3.2 = 71.88 Ω
Final resistance:
R = 230 / 2.8 = 82.14 Ω
Using:
R = R₀ [1 + α (T − 27)]
Substitute:
82.14 = 71.88 [1 + 1.70×10⁻⁴ (T − 27)]
Divide:
1.143 = 1 + 1.70×10⁻⁴ (T − 27)
0.143 = 1.70×10⁻⁴ (T − 27)
T − 27 = 0.143 / (1.70×10⁻⁴)
T − 27 ≈ 841
T ≈ 868°C
Steady temperature ≈ 868°C
Q3.8 A heating element connected to 230 V supply draws 3.2 A initially and 2.8 A when hot. Room temperature is 27°C. (α = 1.70 × 10⁻⁴ °C⁻¹) Find steady temperature.
Initial resistance:
R₀ = V/I₀ = 230 / 3.2 = 71.88 Ω
Final resistance:
R = 230 / 2.8 = 82.14 Ω
Using:
R = R₀ [1 + α (T − 27)]
Substitute:
82.14 = 71.88 [1 + 1.70×10⁻⁴ (T − 27)]
Divide:
1.143 = 1 + 1.70×10⁻⁴ (T − 27)
0.143 = 1.70×10⁻⁴ (T − 27)
T − 27 = 0.143 / (1.70×10⁻⁴)
T − 27 ≈ 841
T ≈ 868°C
Steady temperature ≈ 868°C
Q3.10 (a) In a meter bridge, balance point is 39.5 cm from end A when Y = 12.5 Ω. Find X. (b) Find new balance point if X and Y are interchanged. (c) What happens if galvanometer and cell are interchanged?
(a) In meter bridge:
X/Y = l / (100 − l)
Given:
X/12.5 = 39.5 / 60.5
X = 12.5 × (39.5 / 60.5)
X ≈ 8.16 Ω
(b) On interchanging:
l' = 100 − 39.5 = 60.5 cm
(c) Interchanging galvanometer and cell does not affect balance condition.
X ≈ 8.16 Ω
New balance point = 60.5 cm
Balance condition unaffected
Q3.10 (a) In a meter bridge, balance point is 39.5 cm from end A when Y = 12.5 Ω. Find X. (b) Find new balance point if X and Y are interchanged. (c) What happens if galvanometer and cell are interchanged?
(a) In meter bridge:
X/Y = l / (100 − l)
Given:
X/12.5 = 39.5 / 60.5
X = 12.5 × (39.5 / 60.5)
X ≈ 8.16 Ω
(b) On interchanging:
l' = 100 − 39.5 = 60.5 cm
(c) Interchanging galvanometer and cell does not affect balance condition.
X ≈ 8.16 Ω
New balance point = 60.5 cm
Balance condition unaffected
Q3.12 In a potentiometer, 1.25 V cell balances at 35 cm. If balance point shifts to 63 cm, find emf of second cell.
In potentiometer:
E ∝ l
E₂ / E₁ = l₂ / l₁
E₂ / 1.25 = 63 / 35
E₂ = 1.25 × (63 / 35)
E₂ = 2.25 V
Emf of second cell = 2.25 V
Q3.13 The number density of free electrons in copper is 8.5 × 10²⁸ m⁻³. How long does an electron take to drift across a 3.0 m wire of cross-section 2.0 × 10⁻⁶ m² carrying 3.0 A current?
Given:
n = 8.5 × 10²⁸ m⁻³
A = 2.0 × 10⁻⁶ m²
I = 3.0 A
e = 1.6 × 10⁻¹⁹ C
Length L = 3 m
Using:
I = n e A v_d
v_d = I / (n e A)
Substitute:
v_d = 3 / (8.5×10²⁸ × 1.6×10⁻¹⁹ × 2×10⁻⁶)
v_d ≈ 1.1 × 10⁻⁴ m/s
Time taken:
t = L / v_d
t = 3 / (1.1×10⁻⁴)
t ≈ 2.7 × 10⁴ s
Time ≈ 2.7 × 10⁴ seconds (≈ 7.5 hours)
Q3.14 The earth’s surface has surface charge density 10⁻⁹ C/m². If current flowing to neutralise it is 1800 A, how long will it take? (Radius = 6.37 × 10⁶ m)
Surface area of Earth:
A = 4πR²
A = 4π (6.37×10⁶)²
A ≈ 5.1 × 10¹⁴ m²
Total charge:
Q = σA
Q = 10⁻⁹ × 5.1×10¹⁴
Q = 5.1 × 10⁵ C
Using:
I = Q/t
t = Q/I
t = (5.1×10⁵) / 1800
t ≈ 283 s
Time ≈ 283 seconds (≈ 5 minutes)
Q3.15 (a) Six cells each 2 V, r = 0.015 Ω connected in series to 8.5 Ω resistor. Find current and terminal voltage. (b) A cell of emf 1.9 V and r = 380 Ω. Find maximum current.
(a) Total emf:
E = 6 × 2 = 12 V
Total internal resistance:
r_total = 6 × 0.015 = 0.09 Ω
Total resistance:
R_total = 8.5 + 0.09 = 8.59 Ω
Current:
I = 12 / 8.59
I ≈ 1.4 A
Terminal voltage:
V = E − Ir
V = 12 − (1.4 × 0.09)
V ≈ 11.87 V
(b) Maximum current:
I = E/r
I = 1.9 / 380
I = 0.005 A
Cell cannot drive car motor because very small current.
Current ≈ 1.4 A
Terminal voltage ≈ 11.87 V
Maximum current (old cell) = 0.005 A
Q3.16 Two wires of equal length (Al and Cu) have same resistance. Which is lighter?
Using:
R = ρL/A
For same R and L:
A ∝ ρ
Since ρ_Al > ρ_Cu, aluminium needs larger area.
Mass:
m = density × volume
Even though aluminium has larger area, its density is much smaller.
Density:
Al = 2.7
Cu = 8.9
Hence aluminium wire is lighter.
Aluminium wire is lighter → preferred for overhead cables.
Q3.17 What conclusion can you draw from the following observations on a resistor made of alloy manganin?
From the given table, we observe:
For different values of current and voltage:
V / I ≈ constant
Example:
0.2 A → 3.94 V
R = 3.94 / 0.2 = 19.7 Ω
1.0 A → 19.7 V
R = 19.7 / 1.0 = 19.7 Ω
8.0 A → 158 V
R = 158 / 8 = 19.75 Ω
Thus resistance remains nearly constant.
Conclusions:
1. Ohm’s law is obeyed (V ∝ I).
2. Resistivity of manganin remains nearly constant with temperature.
Manganin obeys Ohm’s law and has very low temperature coefficient of resistance.
Q3.18 Answer the following:
(a) In a non-uniform conductor, which quantity remains constant?
(b) Is Ohm’s law universally applicable?
(c) Why must low voltage supply have small internal resistance?
(d) Why must high tension supply have large internal resistance?
(a) Current remains constant throughout conductor.
(b) No. Ohm’s law is not valid for:
Semiconductors, diodes, transistors, thermistors, vacuum tubes.
(c) For low voltage supply:
I = E / (R + r)
Smaller internal resistance → larger current.
(d) In high tension supply:
Large internal resistance prevents heavy current during short circuit, protecting the system.
(a) Current
(b) Not universal
(c) Small r gives high current
(d) Large r ensures safety
Q3.19 Choose the correct alternative:
(a) Alloys have (greater/less) resistivity than constituent metals.
(b) Alloys have (lower/higher) temperature coefficient.
(c) Manganin resistivity is nearly independent/increases rapidly.
(d) Insulator resistivity is greater than metal by factor 10²² / 10³.
(a) Greater
(b) Lower
(c) Nearly independent
(d) 10²²
Correct answers:
(a) Greater
(b) Lower
(c) Nearly independent
(d) 10²²
Q3.20 (a) For n resistors each R, how to get maximum and minimum resistance?
(b) Combine 1 Ω, 2 Ω, 3 Ω to get (i) 11/3 Ω (ii) 11/5 Ω (iii) 6 Ω (iv) 6/11 Ω.
(a)
Maximum resistance: Series combination
R_max = nR
Minimum resistance: Parallel combination
R_min = R/n
Ratio:
R_max / R_min = n²
(b)
(i) 11/3 Ω → 1 Ω in series with (2 Ω and 3 Ω in parallel)
(ii) 11/5 Ω → 3 Ω in parallel with (1 Ω + 2 Ω)
(iii) 6 Ω → All in series
R = 1 + 2 + 3 = 6 Ω
(iv) 6/11 Ω → All in parallel
1/R = 1 + 1/2 + 1/3 = 11/6
R = 6/11 Ω
Series → Maximum
Parallel → Minimum
Required combinations obtained as above.
Q3.21 Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown. Each resistor has 1 Ω resistance.
Q3.22 In a potentiometer, standard cell 1.02 V balances at 67.3 cm. Unknown cell balances at 82.3 cm. Find emf of unknown cell and answer related conceptual questions.
Q3.23 In potentiometer comparison of resistances, balance length for 10 Ω is 58.3 cm and for X is 68.5 cm. Find X.
Q3.24 In a potentiometer experiment, open circuit balance length is 76.3 cm. With 9.5 Ω resistor, balance length becomes 64.8 cm. Find internal resistance of the cell.