Question

(a) Find the current in the 20 Ω resistor shown in the figure. (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?

Answer 1

(a) Applying Kirchoff's voltage law in loop 1, we get:
In the circuit ABEDA,

10i₁ + 20 (i₁ + i₂) − 5 = 0

⇒ 30i₁ + 20i₂ = 5   ............(1)

Applying Kirchoff's voltage law in loop 2, we get:-

20 (i₁ + i₂) − 5 + 10i₂ = 0

⇒ 20i₁ + 30i₂ = 5  ...........(2)

Multiplying equation (1) by 20 and (2) by 30 and subtracting (2) from (1), we get:-

i₂ = 0.1 A

and i₁ = 0.1 A

∴ Current through the 20 Ω resistor = i₁ + i₂ = 0.1 + 0.1 = 0.2 A

 

(b) Potential drop across across AB is,

\[V_{AB}  = 0 . 2 \times 20 = 4  V\]

Electrostatic energy stored in the capacitor is given by,

\[U = \frac{1}{2}C {V_{AB}}^2 \]

\[U   =   \frac{1}{2} \times 4 \times  {10}^{- 6}  \times (0 . 2 \times 20 )^2 \]

\[U   =   32 \times  {10}^{- 6}   J\]