Question

Consider the circuit shown in figure. Then which of the following statements are true in steady state.

• A. Charge on $6\mu \text{F}$ capacitor is same as on $3\mu \text{F}$ capacitor and is equal to $44\mu \text{C}$
• B. Charge on each capacitor is zero.
• C. Current in ${}^{\prime }{R}^{\prime }$ resistor is zero.
• D. Charge on capacitors depend on ${}^{\prime }{R}^{\prime }$

Answer 1

Answer: (A), (C)

Current in ${}^{\prime }{R}^{\prime }$ resistor is zero.

In steady state, the circuit having capacitors behave as an open circuit. So, the given circuit can be redrawn as shown in the figure below



So, the current passing through this circuit is given by $I=\frac{20-10}{3+2}=2\text{A}$

Since, the current passing through the resistor $R$ is zero (Resistor is a part of open circuit), we can say that the potential difference across the resistor is same. The potential difference across $\text{AB}$ (Shown in figure) is then given by

$V_{AB}=20-(2\times 2)=16\text{V}$


Let us consider charge on $6\mu \text{F}$ capacitor is $q_1$ and on $3\mu \text{F}$ capacitor is $q_2$.

Using $\text{KCL}$ ,$V_A-IR-\frac{q_1}{C_1}+6-\frac{q_2}{C_2}=V_B$

$\Rightarrow 16+6=\frac{q_1}{C_1}+\frac{q_2}{C_2}$

If charge of each capacitor is zero (i.e) $q_1=q_2=0$ then, $\text{L.H.S}\ne \text{R.H.S}$

So, option (b) is wrong.

From the above expression , charges on capacitor plates doesnot have any relation with the resistance $R$.

So, option (d) is also wrong.

If $q_1=q_2=q\left(\text{say}\right)$ then we get, $22=q(\frac{1}{C_1}+\frac{1}{C_2})$

Substituting the data given in the question gives,

$q=22\times (\frac{6\times 3}{6+3})\times {10}^{-6}\text{C}$

$\Rightarrow q=44\mu \text{C}$

Hence, options (a) and (c) are the correct answers.