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 :-

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 :- <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-5prf1qrnaxc1musf.jpg" /> (A) 8× 10- 1J (B) 8× 10- 2J (C) 8× 10- 3J (D) 8× 10- 4J

Tardigrade Admin · Accepted Answer

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 Solution

Applying Kirchoff's first law to the junctions a
and

b,

we find

i_{1} = 3 A

and

i_{2} = 1 A

Now applying Kirchoff's second law to the
closed mash

a e f ba

, 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 (V_{a} - V_{b})^{2} = \frac{1}{2} \times 4 \times 1 0^{- 6} \times (20

= 8 \times 1 0^{- 4} J

Q. 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 :-

$8 \times 1 0^{- 1} J$

$8 \times 1 0^{- 2} J$

$8 \times 1 0^{- 3} J$

$8 \times 1 0^{- 4} J$

Q. 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 :-

8×10−1J

8×10−2J

8×10−3J

8×10−4J

Q. 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 :-

$8 \times 1 0^{- 1} J$

$8 \times 1 0^{- 2} J$

$8 \times 1 0^{- 3} J$

$8 \times 1 0^{- 4} J$