Question

The switch $S$ has been closed for a long time and the electric circuit shown carries a steady current. Let $C_1=3.0\mu \text{F}$, $C_2=6.0\mu \text{F}$, $R_1=4.0\Omega$ and $R_2=7.0\Omega$. The power dissipated in $R_2$ is $2.8\text{kW}$. Then:

• A. The power dissipated in the resistor $R_1$ is $1.6\text{kW}$.
• B. The charge on capacitor $C_1$ is $240\mu \text{C}$.
• C. The charge on capacitor $C_2$ is $440\mu \text{C}$.
• D. When the switch is opened after long time, the charge on $C_1$ is $660\mu \text{C}$.

Answer 1

Answer: (A), (B), (D)

When the switch is opened after long time, the charge on $C_1$ is $660\mu \text{C}$.

In steady state, capacitors in the circuit are open.

So, the circuit diagram become,


Thus, $R_1$ and $R_2$ are in series and $R_{eq}=R_1+R_2=11\Omega$

Since, power dissipated in $R_2=2.8\text{kW}$

Let $i$ be the current passing through both resistors.

$i^2{R}_2=2800\Rightarrow i=20\text{A}$

Therefore, power dissipated in $R_1$ is $i^2{R}_1=(20{)}^2\times 4=1.6\text{kW}$

Using Ohm's law,

$E=i{R}_{eq}=20\times 11=220\text{V}$

Potential drop across $R_1$,

$V_1=i{R}_1=20\times 4=80\text{V}$

So, charge on capacitor $C_1$,

$Q_1=C_1{V}_1=3\times 80=240\mu \text{C}$

Now, potential drop across $R_2$,

$V_2=220-80=140\text{V}$

So, charge on capacitor $C_2$,

$Q_2=C_2{V}_2=6\times 140=840\mu \text{C}$

When switch is opened after long time, $t\to \infty$.

Capacitors behave as conducting wires, therefore both the resistors are in parallel now.

So, the Charge on $C_1$ is

$Q_1^{\prime }=C_{1}E=3\times 220=660\mu \text{C}$

Hence, options $\left(A\right),\left(B\right)$ and $\left(D\right)$ are correct.