Practicing Success
Practicing Success
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The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is 5A, and is decreasing at a rate of 103 A/s then VB – VA = |
15V 20V 25V 30V |
15V |
$\xi= L\frac{dI}{dt} = -5\times 10^{-3}\times 10^3 = -5V$ $V_B + \xi -15 + I\times 1 = V_A$ $V_B - V_A = 15 - \xi - I = 15+5-5 = 15V$ |
