Question

A galvanometer works as 1 V full scale voltmeter
when connected in series with resistance of 2kohm
and as 500 mA ammeter when connected in parallel
with a resistance of 0.2ohm. Find internal resistance
of galvanometer​

Answer 1

internal resistance of galvanometer is 222 Ω

Let internal resistance of galvanometer is G.

galvanometer is connected with R = 2kΩ in series combination.

so, V = (R + G)Ig

⇒Ig = V/(R + G)

here, V = 1 volt so, Ig = 1/(2000 + G) .......(1)

now, an Ammeter is connected in parallel with a resistance of 0.2 Ω

so, Ig × G = (500mA - Ig) × 0.2Ω

⇒Ig × G = (0.5 A - Ig) × 0.2

⇒(0.2 + G) × Ig = 0.1

⇒(0.2 + G) × 1/(2000 + G) = 0.1 [ from eq (1) ]

⇒(0.2 + G) = 200 + 0.1G

⇒0.9G = 199.8

⇒G ≈ 222Ω

hence, internal resistance of galvanometer is 222 Ω

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Answer 2

$\huge\bold\purple{Answer:-}$

Let internal resistance of galvanometer is G.

galvanometer is connected with R = 2kΩ in series combination.

so, V = (R + G)Ig

⇒Ig = V/(R + G)

here, V = 1 volt so, Ig = 1/(2000 + G) .......(1)

now, an Ammeter is connected in parallel with a resistance of 0.2 Ω

so, Ig × G = (500mA - Ig) × 0.2Ω

⇒Ig × G = (0.5 A - Ig) × 0.2

⇒(0.2 + G) × Ig = 0.1

⇒(0.2 + G) × 1/(2000 + G) = 0.1 [ from eq (1) ]

⇒(0.2 + G) = 200 + 0.1G

⇒0.9G = 199.8

⇒G ≈ 222Ω

hence, internal resistance of galvanometer is 222 Ω