Question 172: Electrostatics & Capacitance

Question

Under steady state condition the potential difference across the capacitor in the circuit is $\_\_\_\_$ V.

Accepted Answer

In a DC circuit, at steady state a capacitor is fully charged and it acts as an open circuit i.e., no current flows through the branch containing the capacitor.
The circuit simplifies to a single loop consisting of the 2 V battery, the $6 \Omega$ resistor, and the $2 \Omega$ resistor.
As both the resistors are in series,
$$ \mathrm{R}_{	ext {total }}=6 \Omega+2 \Omega=8 \Omega $$
The steady-state current i flowing through this main loop is:
$$ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{	ext {total }}}=\frac{2}{8}=0.25 \mathrm{~A} $$
The capacitor branch is connected in parallel with the $2 \Omega$ resistor. Therefore, the potential difference across it is equal to the potential difference across the $2 \Omega$ resistor.
$$ \mathrm{V}_{\mathrm{BE}}=\mathrm{i} 	imes 2 \Omega=0.25 	imes 2 \Omega=0.5 \mathrm{~V} $$
The total potential difference $\mathrm{V}_{\mathrm{BE}}$ is equal to that between the capacitor and the $4 \Omega$ resistor. Since there is no current flowing through BE in steady state, the potential drop across the $4 \Omega$ resistor is:
$$ \mathrm{V}_4=\mathrm{i}_{\mathrm{BE}} 	imes 4 \Omega=0 	imes 4=0 \mathrm{~V} $$
Therefore, the entire potential difference of BE is equal to that across the capacitor:
$$ V_C=V_{B E}=0.5 \mathrm{~V} $$
Therefore, the potential difference across the capacitor is 0.5 V .
Thus, the correct option is (A).

Answer

0.5

Answer

1.5

Answer

0

Answer

2

Electrostatics & Capacitance

Under steady state condition the potential difference across the capacitor in the circuit is \_\_\_\_ V.

Choose the correct answer: