A resistance of a galvanometer G=6 Ω maximum current of 2 amp is measured by it. Then required resistance to convert it into an ammeter reading up to 6 A, will be :

Question

A resistance of a galvanometer G=6 Ω maximum current of 2 amp is measured by it. Then required resistance to convert it into an ammeter reading up to 6 A, will be : (A) 5 Ω  (B)  4 Ω  (C)  3 Ω  (D)  2 Ω

Tardigrade Admin · Accepted Answer

Here: Resistance of galvanometer G=6 Ω Maximum current through the galvanometer ig=2A Current in ammeter i=6A The current through galvanometer is given by ig=(S/S +g)i or 2=(S/S+6)× 6 or 25+12=69 So, S=3 Ω

Q. A resistance of a galvanometer $G = 6 Ω$ maximum current of $2 am p$ is measured by it. Then required resistance to convert it into an ammeter reading up to $6 A$ , will be :

$5 Ω$

$4 Ω$

$3 Ω$

$2 Ω$

Here: Resistance of galvanometer
$G = 6Ω$
Maximum current through the galvanometer $i_{g} = 2 A$
Current in ammeter $i = 6 A$
The current through galvanometer is given by
$i_{g} = \frac{S}{S + g} i$
or $2 = \frac{S}{S + 6} \times 6$
or $25 + 12 = 69$
So, $S = 3 Ω$

Q. A resistance of a galvanometer G=6Ω maximum current of 2amp is measured by it. Then required resistance to convert it into an ammeter reading up to 6A, will be :

5Ω

4Ω

3Ω

2Ω