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प्रश्न
(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?
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उत्तर
(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,
\[U = \frac{1}{2}C {V_{AB}}^2 \]
\[U = \frac{1}{2} \times 4 \times {10}^{- 6} \times (0 . 2 \times 20 )^2 \]
\[U = 32 \times {10}^{- 6} J\]
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