A shunt of resistance 1 Ω is connected across a galvanometer of 120 Ω resistance. A current of 5.5 ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly :

Question

A shunt of resistance 1 Ω is connected across a galvanometer of 120 Ω resistance. A current of 5.5 ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly : (A) 5.5 ampere (B) 0.5 ampere (C) 0.004 ampere (D) 0.045 ampere

Tardigrade Admin · Accepted Answer

The current that will given full scale deflection in the absence of the shunt is nearly equal to the current through the galvanometer when shunt is connected i.e.

I_{g}

As

I_{g} = \frac{I S}{G + S}

= \frac{5.5 \times 1}{120 + 1} = 0.045

ampere.

Q. A shunt of resistance 1Ω is connected across a galvanometer of 120Ω resistance. A current of 5.5 ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly :

5.5 ampere

0.5 ampere

0.004 ampere

0.045 ampere

The current that will given full scale deflection in the absence of the shunt is nearly equal to the current through the galvanometer when shunt is connected i.e. Ig​ As Ig​=G+SIS​ =120+15.5×1​=0.045 ampere.

Q. A shunt of resistance $1 Ω$ is connected across a galvanometer of $120 Ω$ resistance. A current of $5.5$ ampere gives full scale deflection in the galvanometer. The current that will give full scale deflection in the absence of the shunt is nearly :

The current that will given full scale deflection in the absence of the shunt is nearly equal to the current through the galvanometer when shunt is connected i.e. $I_{g}$
As $I_{g} = \frac{I S}{G + S}$
$= \frac{5.5 \times 1}{120 + 1} = 0.045$ ampere.