JEE PYQ: Current Electricity Question 43
Question 43 - 2019 (09 Apr 2019 Shift 1)
A moving coil galvanometer has resistance $50,\Omega$ and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a $5,\text{k}\Omega$ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to:
(1) 40 V
(2) 15 V
(3) 20 V
(4) 10 V
Show Answer
Answer: (3)
Solution
$V = I_g(G + R) = 4 \times 10^{-3} \times (50 + 5000) = 4 \times 10^{-3} \times 5050 = 20.2 \approx 20$ V.
BETA
