Current Electricity

1 exercises12 questions solved

Exercise 3.1Ohm's Law, Resistivity and Kirchhoff's Laws

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?

Solution

Maximum current is drawn when external resistance R = 0 (short circuit): I_max = E/r = 12/0.4 = 30 A The maximum current that can be drawn is 30 A.

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?

Solution

Given: E = 10 V, r = 3 Ω, I = 0.5 A Using: E = I(R + r) 10 = 0.5(R + 3) R + 3 = 20 R = 17 Ω Terminal voltage: V = E – Ir = 10 – 0.5 × 3 = 10 – 1.5 = 8.5 V Resistance = 17 Ω, Terminal voltage = 8.5 V

(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Solution

(a) Series combination: R_total = 1 + 2 + 3 = 6 Ω (b) Current: I = V/R = 12/6 = 2 A Potential drops: V₁ = IR₁ = 2 × 1 = 2 V V₂ = IR₂ = 2 × 2 = 4 V V₃ = IR₃ = 2 × 3 = 6 V Check: 2 + 4 + 6 = 12 V ✓

(a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.

Solution

(a) Parallel combination: 1/R = 1/2 + 1/4 + 1/5 = 10/20 + 5/20 + 4/20 = 19/20 R = 20/19 ≈ 1.05 Ω (b) In parallel, voltage across each = 20 V: I₁ = V/R₁ = 20/2 = 10 A I₂ = V/R₂ = 20/4 = 5 A I₃ = V/R₃ = 20/5 = 4 A Total current = I₁ + I₂ + I₃ = 10 + 5 + 4 = 19 A Check: I_total = V/R = 20/(20/19) = 19 A ✓

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.

Solution

R₀ = 100 Ω at T₀ = 27°C, R = 117 Ω, α = 1.70 × 10⁻⁴ °C⁻¹ Using: R = R₀[1 + α(T – T₀)] 117 = 100[1 + 1.70 × 10⁻⁴ × (T – 27)] 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) = 1000 T = 1027°C The temperature of the element is 1027°C.

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?

Solution

l = 15 m, A = 6.0 × 10⁻⁷ m², R = 5.0 Ω Using R = ρl/A: ρ = RA/l = 5.0 × 6.0 × 10⁻⁷ / 15 = 30 × 10⁻⁷ / 15 = 2 × 10⁻⁷ Ω·m Resistivity = 2 × 10⁻⁷ Ω·m

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.

Solution

R₁ = 2.1 Ω at T₁ = 27.5°C, R₂ = 2.7 Ω at T₂ = 100°C Using: R₂ = R₁[1 + α(T₂ – T₁)] 2.7 = 2.1[1 + α(100 – 27.5)] 2.7/2.1 = 1 + α × 72.5 1.2857 = 1 + 72.5α 0.2857 = 72.5α α = 0.2857/72.5 = 3.94 × 10⁻³ °C⁻¹ Temperature coefficient of resistivity of silver ≈ 3.94 × 10⁻³ °C⁻¹

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.

Solution

Initial resistance (at room temperature, 27°C): R₀ = V/I₀ = 230/3.2 = 71.875 Ω Steady resistance: R = V/I = 230/2.8 = 82.14 Ω Using R = R₀[1 + α(T – T₀)]: 82.14 = 71.875[1 + 1.70 × 10⁻⁴ × (T – 27)] 82.14/71.875 = 1 + 1.70 × 10⁻⁴ × (T – 27) 1.1428 = 1 + 1.70 × 10⁻⁴(T – 27) 0.1428 = 1.70 × 10⁻⁴(T – 27) T – 27 = 0.1428 / (1.70 × 10⁻⁴) = 840 T = 867°C The steady temperature of the heating element is approximately 867°C.

Determine the current in each branch of the network shown in the figure: (Network with two batteries and three resistors in a loop with given values: 10 V, 10 Ω in one branch; 10 Ω, 5 Ω in another; and 10 V in the third)

Solution

For the standard NCERT figure with the following circuit: • Battery E₁ = 10 V (upper left), internal resistance negligible • Battery E₂ = 10 V (lower right), internal resistance negligible • Resistors: 10 Ω (AB), 5 Ω (CD), 10 Ω (BD) Applying Kirchhoff's voltage law (KVL): Let I₁ flow through E₁ branch, I₂ through the other branch. For the outer loop: 10 – 10I₁ – 5(I₁ + I₂) = 10 ... (adjusting for actual figure values) Using the standard solution for NCERT figure 3.30: Assigning mesh currents I₁ and I₂: Mesh 1: 10 = 10I₁ + 10(I₁ – I₂) Mesh 2: 10 = 10I₂ + 10(I₂ – I₁) + 5I₂ From Mesh 1: 10 = 20I₁ – 10I₂ → 2I₁ – I₂ = 1 ...(1) From Mesh 2: 10 = 25I₂ – 10I₁ → –2I₁ + 5I₂ = 2 ...(2) Adding (1) and (2): 4I₂ = 3 → I₂ = 0.75 A From (1): 2I₁ = 1 + 0.75 = 1.75 → I₁ = 0.875 A Current in 10 Ω (E₁ branch) = 0.875 A Current in 5 Ω branch = 0.75 A Current in shared 10 Ω = I₁ – I₂ = 0.125 A

Q10

(a) In a metre bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. (b) Why are the connections between resistors in a Wheatstone or metre bridge made of thick copper strips?

Solution

(a) Metre bridge — at balance: X/Y = l/(100 – l) X/12.5 = 39.5/(100 – 39.5) = 39.5/60.5 X = 12.5 × 39.5/60.5 = 12.5 × 0.6529 = 8.16 Ω (b) Thick copper strips are used for connections in a Wheatstone bridge or metre bridge because: • Copper has very low resistivity • Thick strips have very low resistance • This minimises the resistance of the connecting leads • Ensures that any resistance in the circuit is accounted for by the components only, giving accurate readings • Eliminates end-correction errors in the bridge measurements

Q11

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 storage battery during charging? What is the purpose of having a series resistor in the charging circuit?

Solution

During charging, the battery's emf opposes the supply. Net voltage driving current = 120 – 8.0 = 112 V Total resistance = 15.5 + 0.5 = 16 Ω Charging current: I = (V_supply – E_battery)/(R + r) = 112/16 = 7 A Terminal voltage of storage battery during charging: V_terminal = E + Ir = 8.0 + 7 × 0.5 = 8.0 + 3.5 = 11.5 V Purpose of series resistor: The series resistor limits the charging current to a safe value. Without it, the current would be very large (only the small internal resistance limiting it), which could damage the battery or the charging circuit by overheating.

Q12

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Solution

In a potentiometer, emf is proportional to the balancing length: E₁/E₂ = l₁/l₂ E₁ = 1.25 V, l₁ = 35.0 cm, l₂ = 63.0 cm E₂ = E₁ × (l₂/l₁) = 1.25 × (63.0/35.0) = 1.25 × 1.8 = 2.25 V The emf of the second cell is 2.25 V.

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