Two resistances, 6 Ω and 3 Ω, are connected in parallel to an electrical cell of emf 6 V and 1 Ω internal resistance. The reading of the voltmeter connected to the circuit will be

4 V

The correct answer is Option (1) → 4 V

Given:

Resistances: $R_1 = 6\ \Omega$, $R_2 = 3\ \Omega$

Cell emf: $E = 6\ \text{V}$

Internal resistance: $r = 1\ \Omega$

Equivalent resistance of parallel combination:

$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2}$

$R_{\text{eq}} = 2\ \Omega$

Total resistance in circuit: $R_{\text{total}} = R_{\text{eq}} + r = 2 + 1 = 3\ \Omega$

Total current from the cell: $I = \frac{E}{R_{\text{total}}} = \frac{6}{3} = 2\ \text{A}$

Voltage across the parallel combination (also the voltmeter reading):

$V = I \cdot R_{\text{eq}} = 2 \cdot 2 = 4\ \text{V}$

Voltmeter reading = 4 V