Question

Two resistors $400\Omega$ and $800\Omega$ are connected in series with a 6V battery. The potential difference measured by voltmeter of $10\Omega$ across $400\Omega$ resistor is

• A. 2V
• B. 1.95V
• C. 3.8V
• D. 4V

Answer 1

Answer: (B)

Here, the resistances of $400\Omega$ and $10000\Omega$
are in parallel, their effective resistance
$R_P$ will be $R_P=\frac{400\times 10000}{400+10000}$
$=\frac{5000}{13}\Omega$
Total resistance of the circuit,
$=\frac{5000}{13}+800=\frac{15400}{13}\Omega$
Current in the circuit
$I=\frac{6}{15400/13}=\frac{39}{7700}A$
Potential difference across voltmeter
$=I{R}_p=\frac{39}{7700}\times \frac{5000}{13}=1.95V$