In an ammeter, 4 % of the main current is passing through galvanometer. If the galvanometer is shunted with a 5 Ω resistance, the resistance of the galvanometer is:

Question

In an ammeter, 4 % of the main current is passing through galvanometer. If the galvanometer is shunted with a 5 Ω resistance, the resistance of the galvanometer is: (A)  120 Ω  (B)  20 Ω  (C)  5 Ω  (D)  4 Ω

Tardigrade Admin · Accepted Answer

Potential difference across galvanometer resistance and shunt is same.
Let

i_{g}

be the current across galvanometer
and

i - i_{g}

across shunt,
then Potential difference across

G =

potential difference across

S

i.e.,

i_{g} \times G = (i - i_{g}) \times S

Given,

\frac{i _{g}}{i} = \frac{4}{100} = 0.04

,

S = 5 Ω

∴ \frac{G}{5} = \frac{1}{0.04} - 1 = 24

\Rightarrow G = 24 \times 5 = 120 Ω

Q. In an ammeter, $4%$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $5 Ω$ resistance, the resistance of the galvanometer is:

$120 Ω$

$20 Ω$

$5 Ω$

$4 Ω$

Q. In an ammeter, 4% of the main current is passing through galvanometer. If the galvanometer is shunted with a 5Ω resistance, the resistance of the galvanometer is:

120Ω

20Ω

5Ω

4Ω

Q. In an ammeter, $4%$ of the main current is passing through galvanometer. If the galvanometer is shunted with a $5 Ω$ resistance, the resistance of the galvanometer is:

$120 Ω$

$20 Ω$

$5 Ω$

$4 Ω$