Question

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

• A. 2 V
• B. 1.95 V
• C. 3.8 V
• D. 4 V

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
$=i{R}_P$
$=\frac{39}{7700}\times \frac{5000}{13}=1.95V$