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Standard XII

Physics

Resonance in RLC Circuit

The network s...

Question

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is $5$ A and it is decreasing at a rate of $10^{3}$ $A s^{- 1}$ then $V_{B} - V_{A}$ equals.

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Solution

From Kirchhoff law, $V_{A}$ -IR-e- $L \frac{d i}{d t}$ =0
given, I=5A,
e= 15V
L= $5 \times 10^{- 3} H$
$\frac{d i}{d t} = 10^{3} A s^{- 1}$
$V_{A} - I R + e - L \frac{d i}{d t} - V_{B} = 0$
$V_{A} - V_{B} = (5 \times 1) - 15 - (5 \times 10^{- 3}) \times 10^{3} = - 15 V V_{B} - V_{A} = 15 V$

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The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5A and it is decreasing at a rate of 103 As−1 then VB−VA equals.

From Kirchhoff law, VA-IR-e-Ldidt=0given, I=5A,e= 15VL=5×10−3Hdidt=103As−1VA−IR+e−Ldidt−VB=0VA−VB=(5×1)−15−(5×10−3)×103=−15VVB−VA=15V

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is $5$ A and it is decreasing at a rate of $10^{3}$ $A s^{- 1}$ then $V_{B} - V_{A}$ equals.

From Kirchhoff law, $V_{A}$ -IR-e- $L \frac{d i}{d t}$ =0
given, I=5A,
e= 15V
L= $5 \times 10^{- 3} H$
$\frac{d i}{d t} = 10^{3} A s^{- 1}$
$V_{A} - I R + e - L \frac{d i}{d t} - V_{B} = 0$
$V_{A} - V_{B} = (5 \times 1) - 15 - (5 \times 10^{- 3}) \times 10^{3} = - 15 V V_{B} - V_{A} = 15 V$