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A galvanometer having a coil resistance of 60 $\Omega$ shows full scale deflection when a current of 1.0 amp passes through it. It can be converted into an ammeter to read currents upto 5.0 amp by
A
Putting in parallel a resistance of 15 $\Omega$
B
Putting in parallel a resistance of 240 $\Omega$
C
Putting in series a resistance of 15 $\Omega$
D
Putting in series a resistance of 240 $\Omega$
Detailed Solution
A galvanometer is converted into an ammeter by connecting a low resistance (shunt) S in parallel with it.
At full scale, the galvanometer carries $I_g = 1.0$ A and the shunt carries the rest: $I - I_g = 5.0 - 1.0 = 4.0$ A
The potential difference across both is the same: $I_gG = (I - I_g)S$
$S = \frac{I_gG}{I - I_g} = \frac{1.0\times60}{4.0}$
$S = 15\ \Omega$, connected in parallel
At full scale, the galvanometer carries $I_g = 1.0$ A and the shunt carries the rest: $I - I_g = 5.0 - 1.0 = 4.0$ A
The potential difference across both is the same: $I_gG = (I - I_g)S$
$S = \frac{I_gG}{I - I_g} = \frac{1.0\times60}{4.0}$
$S = 15\ \Omega$, connected in parallel