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Byju's Answer

Standard XII

Physics

Work Energy Theorem

In the net wo...

Question

In the net work shown, If at a certain instant the current I is 5 A and decreasing at the rate of $10^{3}$ A/S then $V_{B} - V_{A}$ is
(given R = $1 Ω$ , E = 15 V, L = 5 mH)

5 V

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10 V

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15 V

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20 V

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Solution

The correct option is C 15 V
$V_{A} - i \times R + E - L \frac{d i}{d t} = V_{B}$
$V_{A} - 1 \times 5 + 15 - 5 \times 10^{- 3} \times (10^{3}) = V_{B}$
$V_{B} - V_{A} = - 5 + 15 + 5$
$= 15 V$

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In the net work shown, If at a certain instant the current I is 5 A and decreasing at the rate of 103 A/S then VB−VA is (given R = 1Ω , E = 15 V, L = 5 mH)

The correct option is C 15 VVA−i×R+E−Ldidt=VB VA−1×5+15−5×10−3×(103)=VB VB−VA=−5+15+5 =15 V

In the net work shown, If at a certain instant the current I is 5 A and decreasing at the rate of $10^{3}$ A/S then $V_{B} - V_{A}$ is
(given R = $1 Ω$ , E = 15 V, L = 5 mH)

The correct option is C 15 V
$V_{A} - i \times R + E - L \frac{d i}{d t} = V_{B}$
$V_{A} - 1 \times 5 + 15 - 5 \times 10^{- 3} \times (10^{3}) = V_{B}$
$V_{B} - V_{A} = - 5 + 15 + 5$
$= 15 V$