Question

Find the potential difference across the capacitor in volts

(A) 7
(B) 8
(C) 9
(D) 10

Answer 1

Hint: We firstly need to calculate the equivalent resistance of the circuit by neglecting the capacitor. After calculation of the equivalent resistance of the circuit, we calculate the current in the current following Ohm's law. Further we can calculate the potential at different points and thus the potential difference across the capacitor.

Formula used: V=IR

Complete answer:
Firstly, we neglect the capacitor and calculate the value of equivalent resistance of the entire circuit.

The resistors 1 and 2 are connected in series which gives us an equivalent resistance of 2R. Again the equivalent resistance is in parallel with 3, whose combination gives us a value of 2R/3. The 2 resistors 5 and 6 are in parallel. So the total value of resistance is 5R/3 Ohm.
Now, using Ohm’s law, we calculate the current flowing through the entire circuit
$$\begin{array}{rl}& V=IR\\ & \Rightarrow I={\displaystyle \frac{10}{5R/3}}\\ & \Rightarrow I=6A\end{array}$$
Thus, potential at the left end of the capacitor= (10-2) V
Potential at the right end of the capacitor=0 V

$\therefore$ The potential difference across the capacitor in volts is given as (B)8V.

Additional Information:
The equivalent resistance of the series combination is calculated as:
$R_{eq}=R_1+R_2$, for any two resistors in series having resistance $R_1$ and $R_2$
The equivalent resistance of the parallel combination is calculated as:
${\displaystyle \frac{1}{R_{eq}}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}$ , for any two resistors in parallel having resistance $R_1$ and $R_2$.

Note:
This problem can also be approached with Kirchhoff’s circuit laws for voltage and current after calculation of equivalent voltage of the circuit directly. That might be a more convenient approach for some. But the signs of current must be kept in mind while taking that approach.