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Solved 2019 Question Paper ICSE Class 10 Physics — Question 12

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Question 8(c)

The diagram shows a circuit with the key k open. Calculate: [4]

The diagram shows a circuit with the key k open. Calculate the resistance of the circuit when the key k is open. the current drawn from the cell when the key k is open. the resistance of the circuit when the key k is closed. the current drawn from the cell when the key k is closed. ICSE 2019 Physics Solved Question Paper.

(i) the resistance of the circuit when the key k is open.

(ii) the current drawn from the cell when the key k is open.

(iii) the resistance of the circuit when the key k is closed.

(iv) the current drawn from the cell when the key k is closed.

Answer

(i) When k is open, resistance is 5 + 0.5 = 5.5 Ω

(ii) From relation, V = IR

Substituting the values in the formula we get,

3.3=I×5.5⇒I=3.35.5⇒I=0.6 A3.3 = \text{I} \times 5.5 \\[0.5em] \Rightarrow \text{I} = \dfrac{3.3}{5.5} \\[0.5em] \Rightarrow \text{I} = 0.6 \text{ A}

Hence, current drawn from the cell when the key k is open = 0.6 A.

(iii) When k is closed, the resistance 2 Ω and 3 Ω are in series, therefore their total resistance is 2 Ω + 3 Ω = 5 Ω . Now since, 5 Ω resistance is in parallel to it,

Therefore,

1Rp=15+151Rp=1+15⇒1Rp=25⇒Rp=2.5Ω\dfrac{1}{R_p} = \dfrac{1}{5} + \dfrac{1}{5} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 1}{5} \\[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{2}{5} \\[0.5em] \Rightarrow R_p = 2.5 Ω

So now the resistance of circuit is 2.5 + 0.5 = 3 Ω

(iv) Current drawn I = VR\dfrac{\text{V}}{\text{R}}

Substituting the values we get,

I=3.33⇒I=1.1 A\text{I} = \dfrac{3.3}{3} \\[0.5em] \Rightarrow \text{I} = 1.1 \text{ A}