Electrostatics 6 Question 14
14. Find the potential difference between the points $A$ and $B$ and between the points $B$ and $C$ in the steady state.
(1980)
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Solution:
- Two capacitors of $3 \mu F$ each and two capacitors of $1 \mu F$ each are in parallel.
Therefore, simplified circuit can be drawn as below.
In steady state no current will flow in the circuit and capacitors are fully charged.
Points $A, B$ and $C$ in original circuit are shown in the simplified circuit.
Between points $A$ and $C, 6 \mu F$ and $2 \mu F$ are in series. $100 V$ is applied across this series combination. In series potential drops in inverse ratio of capacity.
$$ \begin{aligned} \therefore \quad V _{A B} & =V _{6 \mu F}=\frac{2}{6+2} \times 100=25 V \\ V _{B C} & =V _{2 \mu F}=\frac{6}{6+2} \times 100=75 V \end{aligned} $$
