JEE PYQ: Current Electricity Question 20
Question 20 - 2020 (03 Sep 2020 Shift 2)
Two resistors $400,\Omega$ and $800,\Omega$ are connected in series across a 6 V battery. The potential difference measured by a voltmeter of $10,\text{k}\Omega$ across $400,\Omega$ resistor is close to:
(1) 2 V
(2) 1.8 V
(3) 2.05 V
(4) 1.95 V
Show Answer
Answer: (4)
Solution
Voltmeter ($10,\text{k}\Omega$) in parallel with $400,\Omega$: $R’ = \frac{10000 \times 400}{10400} = \frac{10000}{26},\Omega$. $I = \frac{6}{\frac{10000}{26} + 800}$. $V = IR’ = \frac{6 \times \frac{10000}{26}}{\frac{10000}{26} + 800} = \frac{150}{77} \approx 1.95$ V.
BETA
