Find the value of current I drawn from the cell.

(a) As, the two 15Ω resistors are in series

So, R_s = 15 + 15 = 30Ω

Now, R_s (i.e., 30Ω) and 30Ω are in parallel,

$$\begin{aligned}\therefore \frac{1}{\text{R}_p} = \frac{1}{30} + \frac{1}{30} \\[1em] = \frac{2}{30} \\[1em] = \frac{1}{15}\end{aligned}$$

∴ R_p = 15Ω

Now, total resistance of the circuit = 2Ω + 15Ω = 17Ω

$$\begin{aligned}\text{I} = \frac{\text{emf}}{\text{total resistance}} \\[1em] = \frac{3.4}{17} \\[1em] = 0.2 \text{ A}\end{aligned}$$

∴ Current I = 0.2 A

(b) Terminal voltage [V] = emf - Ir
= 3.4 - [0.2 x 2] = 3.4 - 0.4
= 3 V

Hence, terminal voltage = 3 V