Two resistors 400Ω and 800Ω are connected in series with a 6 V battery. The potential difference measured by voltmeter of 10 kΩ across 400Ω resistor is:
Answer & explanation
Correct answer: option 2
Here, the resistances of 400Ω and 10000Ω are in parallel, their effective resistance 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 $\mathrm{I}=\frac{6}{15400 / 13}=\frac{39}{7700}$A
Potential difference across voltmeter $=\mathrm{IR}_{\mathrm{P}}$
$=\frac{39}{7700} \times \frac{5000}{13}=1.95 \mathrm{~V}$