The voltage $V_{C_{1}}, V_{C_{2}} a n d V_{C_{3}}$ across the capacitors in the circuit in Fig., under steady state, are respectively.

80 V, 32 V, 48 V

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80 V, 48 V, 32 V

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20 V, 8 V, 12 V

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20 V, 12 V, 8 V

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Solution

The correct option is B 80 V, 48 V, 32 V
At steady state
Inductor behave as short circuit
Capacitor behave as open circuit

$V_{C_{1}} = 100 \times \frac{40}{10 + 40} = 80 V$

$V_{C_{2}} = 80 \times \frac{3}{2 + 3} = 48 V$

$V_{C_{3}} = 80 \times \frac{2}{2 + 3} = 32 V$

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The voltage VC1,VC2andVC3 across the capacitors in the circuit in Fig., under steady state, are respectively.

The correct option is B 80 V, 48 V, 32 VAt steady state Inductor behave as short circuit Capacitor behave as open circuit VC1=100×4010+40=80V VC2=80×32+3=48V VC3=80×22+3=32V

The voltage $V_{C_{1}}, V_{C_{2}} a n d V_{C_{3}}$ across the capacitors in the circuit in Fig., under steady state, are respectively.

The correct option is B 80 V, 48 V, 32 V
At steady state
Inductor behave as short circuit
Capacitor behave as open circuit

$V_{C_{1}} = 100 \times \frac{40}{10 + 40} = 80 V$

$V_{C_{2}} = 80 \times \frac{3}{2 + 3} = 48 V$

$V_{C_{3}} = 80 \times \frac{2}{2 + 3} = 32 V$