Q 12-03-235JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
Two resistors $400\ \Omega$ and $800\ \Omega$ are connected in series across a $6\ \text{V}$ battery. The potential difference measured by a voltmeter of $10\ \text{k}\Omega$ across $400\ \Omega$ resistor is close to:
Answer: (D) $1.95\ \text{V}$
$400\ \Omega$ in parallel with $10\ \text{k}\Omega$: $R_p = \dfrac{400\times10000}{10400} \approx 384.6\ \Omega$.
$$V = 6\times\frac{384.6}{384.6 + 800} \approx 1.95\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics