(a) Find the current in the 20(Omega)resistor shown in figure.(b)If a capacitor of capacitance 4(mu)F is joined between the point A and B,what would be the electrostatic energy stored in it in steady state?

(a) Find the current in the 20(Omega)resistor shown in figure.(b)If a

1.	(a) Find the current in the 20(Omega)resistor shown in figure.(b)If a capacitor of capacitance 4(mu)F is joined between the point A and B,what would be the electrostatic energy stored in it in steady state?
Answer» Solution :Taking circuit ABEDA, ` 10i + 20(i-i_1) - 5 = 0 ` ` rArr10i + 20I - 20i -5 =0 ` ` rArr 30i - 20i_1 -5 = 0 ` Taking circuit abfca, ` 20(i-i_1)- 5 - 10i_1 = 0 ` ` rArr 20i -10i_1 - 20i_1 - 5 =0 ` ` rArr 20i - 30i_1 -5 =0 ` ` (30i-20i_1 -5 =0 )2 ` ` (20i - 30i_1-5 =0 )3 ` ` (90-40)i_1 =0 ` ` rArr i_1 = 0 ` `rArr30i - 5 = 0 ` ` i = 5/30 = 0.16A` ` CURRENT through 20 Omega is 0.16A. `

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(a) Find the current in the 20(Omega)resistor shown in figure.(b)If a capacitor of capacitance 4(mu)F is joined between the point A and B,what would be the electrostatic energy stored in it in steady state?

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