A switch is briefly closed to connect a 30Volt battery to a 670mH inductor in pa

Question

A switch is briefly closed to connect a 30Volt battery to a 670mH inductor in parallel with a 4 resistor.

At the instant the switch is closed, there is already a current i = 2 A flowing through the inductor. If this current is increased while the switch remains closed for 14 ms, find:

1. The amount of energy delivered to the inductor by the battery while the switch is closed ______mJ.

2.   The amount of power being delivered by the battery at the end of this process (just before the switch is re-opened) ____________Watts.

Explanation / Answer

1.
V = L*dI/dt
so, 30 = .67*dI/.014
so, dI = 0.627 A
so, energy delivered = .5*.67*(2.627^2 -2^2) = 971.8 mJ

2. power delivered = 971.8/.014 = 69.4 watts

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.