Electrostatics Ques 150
- 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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Answer:
Correct Answer: 150.$V_{AB}=25 V, V_{BC}=75 V$
Solution:
Formula:
Charging and Discharging of a Capacitor:
- Two capacitors of $3 \mu \mathrm{F}$ each and two capacitors of $1 \mu \mathrm{F}$ 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 $C, 6 \mu \mathrm{F}$ and $2 \mu \mathrm{F}$ are in series. $100 \mathrm{~V}$ is applied across this series combination. In series potential drops in inverse ratio of capacity.
$$ \begin{aligned} \therefore \quad V_{A B} & =V_{6 \mu \mathrm{F}}=\Big(\frac{2}{6+2}\Big) \times 100=25 \mathrm{~V} \\ V_{B C} & =V_{2 \mu \mathrm{F}}=\Big(\frac{6}{6+2}\Big) \times 100=75 \mathrm{~V} \end{aligned} $$
BETA
