Question

A part of circuit in steady state along with the currents flowing in the branches, the value of resistances etc., is shown in figure. Calculate the
energy stored in the capacitor :-

• A. $8\times 10^{- 1}J$
• B. $8\times 10^{- 2}J$
• C. $8\times 10^{- 3}J$
• D. $8\times 10^{- 4}J$

Answer 1

Answer: (D)

When the capacitor plates get fully charged, there
will be no current in branch $ab$ . Remember
capacitance acts as the open circuit since
capacitance offers infinite resistance to $d.c.$ The
capacitance simply collects the charge Applying Kirchoff's first law to the junctions a
and $b,$ we find $i_{1}=3A$ and $i_{2}=1A$
Now applying Kirchoff's second law to the
closed mash $aefba$ , we get
$3\times 5+3\times 1+1\times 2=V_{a}-V_{b}$
or $3 \times 5+3 \times 1+1 \times 2= V _{ a }- V _{ b }$
or $V _{ a }- V _{ b }=20 V$
$U =\frac{1}{2} C \left( V _{ a }- V _{ b }\right)^2=\frac{1}{2} \times 4 \times 10^{-6} \times(20$
$=8 \times 10^{-4} J$