Question 18: Semiconductors and Logic Gates

Question

The charge stored by the capacitor C in the given circuit in the steady state is __________ µC.

Accepted Answer

At steady state (very long time), in a DC circuit, a capacitor is fully charged and it acts as an open circuit. No current flows through the branch containing the capacitor.
The diodes are ideal which acts as a closed switch (conducting) in forward-biased state and acts as a open switch (non-conducting) in reverse-biased state.
$1 \Omega$ resistor is connected directly across the battery. So, the current flows here.
$2 \Omega$ resistor is connected directly across the battery. Current flows here before reaching steady state.
The diode $D_1$ is pointing toward the positive terminal. This means it is reverse-biased. So, no current flows through this branch.
The diode $\mathrm{D}_2$ is pointing away from the positive terminal toward the negative terminal. This means it is forward-biased. Current flows through this branch.
In the steady state, when capacitor is fully charged and acts as open circuit; the resulting circuit will be,
The voltage across the capacitor is equal to the voltage across points B and G , i.e., between A and H .
The current in the close loop ABCDEFGH ,
$$ \mathrm{i}=\frac{2.5}{1+4} \mathrm{~A}=0.5 \mathrm{~A} $$
So, the voltage across the capacitor is, $\mathrm{V}_{\mathrm{BG}}=\mathrm{V}_{\mathrm{AH}}=(2.5-1 	imes 0.5) \mathrm{V}=2.0 \mathrm{~V}$
The formula for charge stored in a capacitor is :
$$ \mathrm{Q}=\mathrm{C} 	imes \mathrm{V} $$
The capacitance is $\mathrm{C}=5 \mu \mathrm{~F}$
$$ Q=5 \mu \mathrm{~F} 	imes 2.0 \mathrm{~V} $$
$$\Rightarrow $$ $$ Q=10.0 \mu \mathrm{C} $$
Therefore, the charge stored by the capacitor in the steady state is $10.0 \mu \mathrm{C}$.
Hence, the correct option is (C).

Answer

7.5

Answer

5

Answer

10

Answer

12.5

Semiconductors and Logic Gates

The charge stored by the capacitor C in the given circuit in the steady state is __________ µC.

Choose the correct answer: