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The network shown in the figure is part of a completecircuit. Ifat a certain instant, the current I is 5A and it isdecreasing at a rate of 103 As ' then V-V equals12+ 5 mH15 V(A) 20 V(C) 10 V(B) 15 V(D)5 VIn previous problem if I is reversed in direction, thenVg-V equals(A)5V(C) 15 V(B) 10 V(D) 20 V |
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Answer» the voltage drop El = -ldi/dt = -5*(10^-3)*(10³) =-5v so, the voltage drop Vb= Va -5+15-5 => Vb-Va = 5V |
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