Question

(a) Find the current in the 20 Ω resistor shown in the figure (32-E31). (b) If a capacitor of capacitance 4 μF is joined between the points A and B, what would be the electrostatic energy stored in it in steady state?
Figure

Answer 1

Explanation:

(a) In loop 1, when Kirchoff’s voltage law is applied, we obtain:

In the ABEDA circuit,

20 (i1 + i2) + 10i1 − 5 = 0

⇒ + 20i2 + 30i1 = 5    …(i)

In loop 2, applying Kirchoff’s voltage law, we obtain:

10i2 + 20 (i1 + i2) − 5 = 0

⇒ 20i1 + 30i2 = 5    …(ii)

When equation (i) is divided by 20 and equation (ii) is divided by 30 and equation (ii) subtracted from equation (i), we obtain:

i2 = 0.1 A

and i1 = 0.1 A

Therefore, the current passing via the 20 Ω resistor = i2 + i1  = 0.1 + 0.1 = 0.2 A

(b) Across AB, the potential drop is,

VAB = 0.2 $\times$ 20 = 4 V

In the capacitor, the electrostatic energy stored is shown as,

U = 12CVAB2U = 12 $\times$ 4 $\times$ 10-6 $\times$ (0.2$\times$20)2U = 32 $\times$ 10-6 J