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Solved 2020 Question Paper ICSE Class 10 Physics — Question 12

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Question

Question 8(c)

The diagram alongside shows three resistors connected across a cell of e.m.f. 1.8 V and internal resistance r. Calculate: [4]

The diagram alongside shows three resistors connected across a cell of e.m.f. 1.8 V and internal resistance r. Calculate Current through 3 Ω resistor and internal resistance r. ICSE 2020 Physics Solved Question Paper.

(i) Current through 3 Ω resistor.

(ii) The internal resistance r.

Answer

(i) The equivalent resistance Rp of resistances 3 Ω and 1.5 Ω joined in parallel is given by

1Rp=13+23  [∵1.5Ω=32Ω]⇒1Rp=33⇒Rp=1Ω\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{2}{3} \space \space \big[\because 1.5 Ω = \dfrac{3}{2} Ω\big] \\[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{3}{3} \\[0.5em] \Rightarrow R_p = 1 Ω

From, V = IR

the potential drop across this combination = 0.3 x 1 = 0.3 V

∴Current through 3 Ω resistance=VoltageResistance=0.33=0.1 A\therefore \text{Current through 3 Ω resistance} = \dfrac{\text{Voltage}}{\text{Resistance}} \\[0.5em] = \dfrac{0.3}{3} = 0.1 \text{ A}

(ii) Total resistance of the circuit = Rp + 4 + r = (5 + r) Ω

Current in the circuit,

0.3=1.85+r0.3×(5+r)=1.8(0.3×5)+(0.3×r)=1.81.5+(0.3×r)=1.80.3×r=1.8−1.5r=0.30.3=1Ω0.3 = \dfrac{1.8}{5 + r} \\[0.5em] 0.3 \times (5 + r) = 1.8 \\[0.5em] (0.3 \times 5) + (0.3 \times r) = 1.8 \\[0.5em] 1.5 + (0.3 \times r) = 1.8 \\[0.5em] 0.3 \times r = 1.8 - 1.5 \\[0.5em] r = \dfrac{0.3}{0.3} = 1 Ω

∴ Internal resistance r = 1 Ω