Physics 2004 JAMB Past Questions - Question 41: A steady current of 2A flows in a coil of emf 12V for 0.4s. A back emf of 3V was induced during this period. The stored energy in the loop that can be utilized is

Correct Answer

C

Explanation

To find the stored energy in the coil that can be utilized, you can use the formula for the energy stored in an inductor:

Energy (W) = 1/2 * L * I^2

Where:
W = Energy stored in the inductor (in joules)
L = Inductance of the coil (in henrys)
I = Current flowing through the coil (in amperes)

In your case, you have a coil with an emf (voltage) of 12V, a current of 2A, and a back emf of 3V during a period of 0.4 seconds.

First, we need to find the inductance (L) of the coil. We can use the formula for back emf in an inductor:

Back EMF (E) = L * dI/dt

Where:
E = Back EMF (in volts)
L = Inductance (in henrys)
dI/dt = Rate of change of current (in amperes per second)

Given that the back emf is 3V and the current change occurred in 0.4 seconds, we can rearrange the formula to find L:

L = E / (dI/dt)

L = 3V / (2A / 0.4s) = 6 H (henrys)

Now that we have the inductance (L), we can calculate the energy stored in the coil:

W = 1/2 * 6H * (2A)^2 = 1/2 * 6H * 4A^2 = 12 joules

So, the stored energy in the coil that can be utilized is 12 joules.