Question

A galvanometer of resistance 100 ohms is shunted so that only \[\dfrac{1}{{11}}\] of the main current flows through the galvanometer. The resistance of the shunt is
A. 1 ohm
B. 11 ohms
C. 10 ohms
D. 9 ohms

Answer 1

Hint: A shunt is a low value resistor that is connected in parallel with the galvanometer to decrease the net resistance of the combination so that the combination can behave as an ammeter (a device used to measure the current).

Complete step by step answer:
A galvanometer is a device used to detect the current in the electric circuit. A galvanometer can be converted into an ammeter by connecting a low resistance shunt in parallel with the galvanometer.
The value of shunt is given by $S = \dfrac{{\mathop I\nolimits_g G}}{{I - \mathop I\nolimits_g }}$ …(1)
Where $I = $ Current through the whole circuit
 $\mathop I\nolimits_g = $ Current through the galvanometer
$G = $ Resistance of galvanometer
$S = $ Resistance of shunt
We have a galvanometer of resistance 100Ω and only $\dfrac{1}{{11}}$ of the main current flows through the galvanometer. Let the current flowing through the main circuit be $I$.
Now we have,
$G = 100\Omega $ and $\mathop I\nolimits_g = \dfrac{1}{{11}} \times I = \dfrac{I}{{11}}$
Putting these values in (1) we get
$S = \dfrac{{\dfrac{I}{{11}} \times 100}}{{I - \dfrac{I}{{11}}}} = \dfrac{{\dfrac{{100I}}{{11}}}}{{\dfrac{{11I - I}}{{11}}}} = \dfrac{{100I}}{{10I}} = 10\Omega $
So the resistance of the shunt is 10Ω.

So, the correct answer is “Option C”.

Additional Information:
This question can also be solved by another method as follows:
Let the main current be $I$ and the resistance of shunt be $S$ then the current through galvanometer will be $\dfrac{I}{{11}}$ and we know that the resistance of galvanometer is 100Ω.
In a parallel combination, the voltage at both the resistors is always equal.
So Voltage at galvanometer = Voltage at shunt and we know voltage = current $\times$ resistance
∴ $
 \dfrac{I}{{11}} \times 100 = \dfrac{{10I}}{{11}} \times S \\
 S = 10\Omega \\
 $

Note:
Net resistance in a parallel combination is lower than the smallest resistance present in the circuit that’s why a shunt is connected in parallel with the galvanometer to convert it into an ammeter. In a parallel combination, voltage across the resistors remains equal while the current divides between them based on their resistances.