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

(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,

$$\begin{aligned}3.3 = \text{I} \times 5.5 \\[0.5em] \Rightarrow \text{I} = \frac{3.3}{5.5} \\[0.5em] \Rightarrow \text{I} = 0.6 \text{ A}\end{aligned}$$

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,

$$\begin{aligned}\frac{1}{R_p} = \frac{1}{5} + \frac{1}{5} \\[0.5em] \frac{1}{R_p} = \frac{1 + 1}{5} \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{2}{5} \\[0.5em] \Rightarrow R_p = 2.5 Ω\end{aligned}$$

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

(iv) Current drawn I = $\frac{\text{V}}{\text{R}}$

Substituting the values we get,

$$\begin{aligned}\text{I} = \frac{3.3}{3} \\[0.5em] \Rightarrow \text{I} = 1.1 \text{ A}\end{aligned}$$