Current Electricity 4 Question 6
6. The galvanometer deflection, when key $K _1$ is closed but $K _2$ is open equals $\theta _0$ (see figure). On closing $K _2$ also and adjusting $R _2$ to $5 \Omega$, the deflection in galvanometer becomes $\frac{\theta _0}{5}$. The resistance of the galvanometer is given by (neglect the internal resistance of battery) :
(Main 2019, 12 Jan I)
(a) $22 \Omega$
(b) $5 \Omega$
(c) $25 \Omega$
(d) $12 \Omega$
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Solution:
- For a galvanometer, $i _g \propto \theta$ or $i _g=C \theta$
where, $I _g=$ current through coil of galvanometer,
$\theta$ = angle of deflection of coil
and $C$ is the constant of proportionality.
Now, $K _1$ is closed and $K _2$ is opened, circuit resistance is
$$ R _{eq}=R _1+R _g $$
where, $R _g=$ galvanometer resistance.
Hence, galvanometer current is
i.e.,
$$ i _g=\frac{V}{R _1+R _g} $$
where, $V=$ supply voltage.
As deflection is given $\theta _0$, we have
$$ i _g=\frac{V}{R _1+R _g}=C \theta _0 $$
When both keys $K _1$ and $K _2$ are closed, circuit resistance.
$$ R _{eq}=R _1+\frac{R _2 \times R _g}{R _2+R _g} $$
Current through galvanometer will be
$$ \begin{aligned} i _{g _2} & =\frac{V}{R _1+\frac{R _2 R _g}{R _2+R _g}} \times \frac{R _2}{\left(R _2+R _g\right)} \\ & =\frac{V R _2}{R _1 R _2+R _1 R _g+R _2 R _g}=C \cdot \frac{\theta _0}{5} \end{aligned} $$
Now, dividing Eq. (i) by Eq. (ii), we get
$$ \frac{R _1 R _2+R _1 R _g+R _2 R _g}{R _2\left(R _1+R _g\right)}=5 $$
Substituting $R _1=220 \Omega$ and $R _2=5 \Omega$,
we get
$$ \begin{aligned} \frac{1100+220 R _g+5 R _g}{5\left(220+R _g\right)} & =5 \\ 1100+225 R _g & =5500+25 R _g \\ \Rightarrow \quad 200 R _g & =4400 \end{aligned} $$
$$ \begin{array}{ll} \Rightarrow & R _g=\frac{4400}{200}=22 \Omega \\ \therefore & R _g=22 \Omega \end{array} $$
