Question

A resistor with r = 20.0 Ω and an inductor with l = 0.600 h are connected in series to a battery that has emf 70.0 v and negligible internal resistance. At time t after the circuit is completed, the energy stored in the inductor is ______ j.

Answer 1

The energy stored in the inductor is approximately 3.668 J.

Step-by-step explanation:

The energy stored in an inductor is given by the formula:

UL = (1/2) * L * I^2

where UL is the energy stored in the inductor, L is the inductance, and I is the current.

In the given question, the inductance (L) is 0.600 H. To find the current (I), we need to calculate the total resistance in the circuit which is the sum of the resistance of the resistor (R) and the inductor (r). Therefore, the total resistance (RT) is 20.0 Ω + 0.600 Ω = 20.6 Ω. We can now use Ohm's Law to find the current:

I = V / RT

substituting the given emf and resistance:

I = 70.0 V / 20.6 Ω ≈ 3.398 A

Now we can calculate the energy stored in the inductor:

UL = (1/2) * L * I^2

UL ≈ (1/2) * 0.600 H * (3.398 A)^2 ≈ 3.668 J