Question

# In the circuit shown, initially there is no charge on capacitors and keys ${{S}_{1}}$ and ${{S}_{2}}$ are open. The values of the capacitors are ${{C}_{1}}=10\mu F$, ${{C}_{2}}=10\mu F$, and ${{C}_{3}}={{C}_{4}}=80\mu F$. Which statement is/are correct?A.) The key ${{S}_{1}}$ is kept closed for a long time such that the capacitors are fully charged. Now the key ${{S}_{2}}$ is closed, at this time the instantaneous current across $3\Omega$ resistor (between points P and Q) will be $0.2A$ (rounded off to first decimal place).B.) If the key ${{S}_{1}}$ is kept closed for a long time such that capacitors are fully charged, the voltage across ${{C}_{1}}$ will be $4V$.C.) At time $t=0$, the key ${{S}_{1}}$ is closed, the instantaneous current in the closed circuit will be $25mA$.D.) If ${{S}_{1}}$ is kept closed for a long time such that capacitors are fully charged, the voltage difference between P and Q will be $10V$.

Hint: In a steady state circuit, the voltage across the capacitor becomes constant and the capacitor acts like an open circuit. We will use this concept and apply the equation $q=CV$ to find charge on the capacitor plates. For finding instantaneous current, we will use KVL in different loops.

The moment we close the switch ${{S}_{1}}$, charge on the capacitor becomes zero. All capacitors can be replaced by a wire. Values of capacitors are $\text{70 }\!\!\Omega\!\!\text{ , 30 }\!\!\Omega\!\!\text{ , and 100 }\!\!\Omega\!\!\text{ }$. After closing key ${{S}_{1}}$, voltage in the circuit is $5V$ across the points P and Q. The current flowing in the circuit will be$i=\dfrac{V}{70+100+30}=\dfrac{5}{200}=25$

We get, $i=25mA$

Now, if key ${{S}_{1}}$ is kept closed for a long time, the capacitor will be in its steady state because of continuous flow of charge in the circuit.

For a capacitor, we will use the formula $q=CV$ where $q$ is the charge on the capacitor, $C$ is the capacitance, and $V$ is the voltage drop across the capacitor.

Applying KVL in the loop for which ${{S}_{1}}$ is closed,

\begin{align} & \dfrac{q}{{{C}_{1}}}+\dfrac{q}{{{C}_{2}}}+\dfrac{q}{{{C}_{3}}}-5=0 \\ & \dfrac{q}{10}+\dfrac{q}{80}+\dfrac{q}{80}-5=0 \\ & \dfrac{10q}{80}=5 \\ & q=40 \\ \end{align}

Voltage across ${{C}_{1}}=\dfrac{q}{{{C}_{1}}}=\dfrac{40}{10}=4V$

We get the value of $q$ equals to $40\mu C$.

Now the moment the key ${{S}_{2}}$ is closed, the charge on the capacitor will remain the same.
Applying KVL again we get
\begin{align} & (-10+x)\left( 30+\dfrac{40}{10}+70y \right)=0 \\ & 30x+70y-6=0 \\ \end{align}

Let $x$ be the instantaneous current in the circuit while key ${{S}_{1}}$ is closed.

Voltage drop in the circuit $=5V$ and equivalent resistance will be the sum of resistances $\text{70 }\!\!\Omega\!\!\text{ , 30 }\!\!\Omega\!\!\text{ , and 100 }\!\!\Omega\!\!\text{ }$, as all three are present in series in the circuit.
We get $i=\dfrac{V}{R}=\dfrac{5}{200}=0.025$

$i=25mA$

After closing key ${{S}_{2}}$, we can find the value of current flowing in the branch PQ.
Equivalent resistance in this case will be $151\Omega$

\begin{align} & {{I}_{PQ}}=\dfrac{6\times 2}{151}=0.08 \\ & {{I}_{PQ}}=80mA \\ \end{align}

If the key ${{S}_{1}}$ is kept closed for a long time such that capacitors are fully charged, the voltage across ${{C}_{1}}$ will be $4V$. Also, at time $t=0$, the key ${{S}_{1}}$ is closed, the instantaneous current in the closed circuit will be $25mA$.

Hence, correct options are B and C.

Note: Students should keep in mind the concept of steady state. Steady state can be achieved by keeping the switch opened or closed for a long period of time. There must be no accumulation of mass or energy over the steady state time and the system is present in total equilibrium. Apply all the equations very carefully.