Question

The figure shown is a part of a circuit at steady state. The current flowing from $3\Omega$ resistor is :

• A. $2A$
• B. $\frac{4}{3}A$
• C. $\frac{2}{3}C$
• D. None of these

Answer 1

The correct option is B $\frac{4}{3}A$

In steady state, capacitors behave like open circuit ( i.e. no current will pass through capacitors).

So $2\Omega$ and $4\Omega$ will be in series, and resultant

resistance $R_1$ will be parallel to$3\Omega$ , and new resultant

resistance $R_2$ will be in series with $4\Omega$ lets call

it $R_3$, now applying junction law at point P , $R_3$= $6\Omega$

$(4-p)/2+(3-p)/1+(-10-p)/R_3=0$

we get, $=2v$

Similarly finding potential of other end Q of $3\Omega$, by junction law

$(2-Q)/2+(-10-Q)/4=0$

we get $Q=-2V$, so current through $3\Omega$ is

$[2-(-2\left)\right]V/3\Omega =4/3A$