The below diagram shows the part of a complete circuit. The ammeter reads 1A.

Given,

• Current through the 6Ω resistor = ammeter reading = 1 A

(a) The 6Ω and 3Ω resistors are joined in parallel, so the potential difference across both of them is the same.

$$\begin{aligned}1 \times 6 = \text I_1 \times 3 \\[1em] \Rightarrow \text I_1 = \frac{6}{3} = 2 \text { A}\end{aligned}$$

Hence, the current through the 3Ω resistor is 2 A.

(b) In a parallel combination the main current is the sum of the currents in the branches.

I = 1 + I₁ = 1 + 2 = 3 A

Hence, the total current in the circuit is 3 A.

(c) False.

Reason — When the switch S is opened, the 3Ω branch is broken and only the 6Ω resistor is left in the circuit. The resistance of the parallel combination was

$\text R_\text p = \frac{6 \times 3}{6 + 3} = 2\ \Omega$

which now rises to 6 Ω. For the same supply voltage, an increase in resistance decreases the current.

Hence, the ammeter reading will become less than 1 A, and not (1 + I₁) = 3 A.