Question

Consider the circuit shown in the figure where a battery of e.m.f. $4\text{V}$ and a capacitor of capacitance $1\mu \text{F}$ is connected to a combination of resistances. Find out the steady-state current in the battery.

• A. $1.5\text{A}$
  
• B. $2\text{A}$
  
• C. $2.5\text{A}$
  
• D. $3.5\text{A}$
  

Answer 1

The correct option is B $2\text{A}$


Let us consider $I$ to be the steady state current through the circuit.

In the steady-state condition, current $I$ is constant in the circuit and the capacitors offer infinite resistance.

So, the resistance $10\Omega$ becomes ineffective in the circuit. So, in this case, the equivalent resistance of resistors $2\Omega$ and $4\Omega$ that are connected in parallel is,

$\frac{1}{R_p}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

$\Rightarrow R_p=\frac{4}{3}\Omega$

Therefore, the total resistance of the circuit$=\frac{4}{3}+\frac{2}{3}=\frac{6}{3}=2\Omega$

Hence, the steady-state current in the circuit , $I=\frac{4}{2}=2\text{A}$

Hence option (b) is the correct answer.