Question

A galvanometer is connected as shown in figure. It has resistance of $100\Omega$. What should be resistance connected to it in parallel so that its deflection is reduced to half of the initial value ?

• A. $100\Omega$
• B. $91\Omega$
• C. $99\Omega$
• D. None of these

Answer 1

The correct option is B $91\Omega$

Initially when there was no resistance in parallel the circuit was

Total resistance of circuit

$100+1000=1100\Omega$

Voltage, $V=2V$

$⟹I=\frac{V}{R}=\frac{2}{1100}=1.8\times {10}^{-}3A$

$⟹$ When a current of $1.8\times {10}^{-}3A$ flows through circuit it shows full deflection $I_g=1.8\times {10}^{-}3A$

When $⟹I_g=\frac{1.8\times {10}^{-}3A}{2}=0.9\times {10}^{-}3A$

It will show half deflection.

$⟹I-I_1=0.9\times {10}^{-}3A\to \left(1\right)$

Using Kirchhoff's law for loop $ABCDEFGHA$

$S\left(I_1\right)+I\left(1000\right)-2=0\to \left(2\right)$

For loop $ABCFGHA$

$\left(100\right)(I-I_1)+I\left(1000\right)-2=0\to \left(3\right)$

Substituting $\left(1\right)$ in $\left(3\right)$ we get,

$100(0.9\times {10}^{-3})+I\left(1000\right)-2=0$

$0.9\times {10}^{-}3+1000I-2=0$

$1000I=2-0.09=1.91$

$I=\frac{1.91}{1000}=1.91\times {10}^{-3}\to \left(4\right)$

Substituting $\left(4\right)$ in $\left(1\right)$ we get

$I_1=1.01\times {10}^{-3}A\to \left(5\right)$

Substituting equation $\left(5\right),\left(4\right)$ in $\left(2\right)$ we get

$S(1.01\times {10}^{-3})+1.91-2=0$

$s(1.01\times {10}^{-3}=0.09)$

$S=\frac{0.09}{1.01\times {10}^{-3}}=\frac{9000}{101}$

$\approx 91\Omega$