Q. Two resistors $400\ \Omega$ and $800\ \Omega$ are connected in series with 6 V battery. The potential difference measured by voltmeter of $10k\ \Omega$ across $400\ \Omega$ resistor is
A) 2 V
B) 1.95 V
C) 3.8 V
D) 4 V
Explanation
(b) Here, the resistance of 400$\ \Omega\ $and 10000$\ \Omega$ are in parallel, their effective resistance $R_{p}\ $will be
$R_{P} = \frac{400 \times 10,000}{400 + 10,000} = \frac{5000}{13}\Omega$
Total resistance of 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=$iR_{P} = \frac{39}{7700} \times \frac{5000}{13}$
$= 1.95\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.