A balanced bridge is shown in the circuit diagram. The metre bridge wire has resistance $1 \Omega…
A balanced bridge is shown in the circuit diagram. The metre bridge wire has resistance $1 \Omega \mathrm{cm}^{-1}$. The current drawn from the battery is (Internal resistance of battery is negligible)
$0.44 \mathrm{~A}$
$0.66 \mathrm{~A}$
$0.88 \mathrm{~A}$
$0.22 \mathrm{~A}$
Solution
Wire has a resistance $1 \Omega \mathrm{cm}^{-1}$.
$\therefore$ Resistance of $40 \mathrm{~cm}$ of wire $=40 \Omega$
and resistance of $60 \mathrm{~cm}$ of wire $=60 \Omega$
No current flows through the galvanometer since the bridge in balanced. Hence the branch containing the galvanometer can be removed.
For balanced bridge $\frac{4}{40}=\frac{Y}{60}$
$\therefore \mathrm{Y}=\frac{4 \times 60}{40}=6 \Omega$
$4 \Omega$ and $6 \Omega$ are in series. Their equivalent resistance is $10 \Omega$. $40 \Omega$ and $60 \Omega$ are in series. Their equivalent resistance is $100 \Omega$. $10 \Omega$ and $100 \Omega$ are in parallel Their equivalent resistance is $\frac{1000}{110} \Omega$.
$\therefore$ The current $I=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \times 110}{1000}=0.66 \mathrm{~A}$
