Question

Ideal voltmeter connected as shown reads

• A. 16 V
• B. 12 V
• C. 4 V
• D. 8 V

Answer 1

Answer: (D)

8 V

Answer 2

Since $6 \Omega$ and $10 \Omega$ are in parallel so its equivalent resistance,
$R = \frac{6 \times 12}{6 + 12} = \frac{6 \times 12}{18} = 4 \Omega$
Net resistance of the circuit , $R_{eq} = 4 \Omega + 4 \Omega$
\hspace60mm = 8 \Omega
Current in the circuit, $I = \frac{V}{R_{eq}} = \frac{16}{8} = 2 A.$
Reading of voltmeter= voltage drop across $4 \Omega$
$= I \times R=2 \times 4=8V$