JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 17)
A moving coil galvanometer has resistance 50$$\Omega $$
and it indicates full deflection at 4mA current.
A voltmeter is made using this galvanometer
and a 5 k$$\Omega $$ resistance. The maximum voltage,
that can be measured using this voltmeter, will
be close to :
15 V
10 V
40 V
20 V
Explanation
G = 50 $$\Omega $$
S = 5000 $$\Omega $$
Ig = 4 × 10–3
V = ig (G + S)
V = 4 × 10–3 (50 + 5000)
= 4 × 10–3 (5050) = 20.2 volt
S = 5000 $$\Omega $$
Ig = 4 × 10–3
V = ig (G + S)
V = 4 × 10–3 (50 + 5000)
= 4 × 10–3 (5050) = 20.2 volt
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