Find the current in the 20Ω resistor shown in figure (32-E31). (b) If a capacitor of capacitance 4 μF is joined between the points A and B,

Question

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

Answer 1

 Applying Kirchoff’s voltage law in loop 1, we get:
In the circuit ABEDA,
10i₁ + 20 (i₁ + i₂) − 5 = 0
⇒ 30i₁ + 20i₂ = 5    …(i)

Applying Kirchoff’s voltage law in loop 2, we get:
20 (i₁ + i₂) − 5 + 10i₂ = 0
⇒ 20i₁ + 30i₂ = 5    …(ii)
Multiplying equation (i) by 20 and (ii) by 30 and subtracting (ii) from (i), 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

Comments

i think the method is wrong .

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corrected

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It's correct
Apply loop law in the direction of current for any one of the small loop and the bigger one

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For b) v=ir
              =0.2*20 = 4v
      Energy stored =1/2 cv2
                                  = 32muj