A coil has an inductance of \(0.15 \mathrm{H}\) and a resistance of 33 \(\Omega\). The coil is connected to a 6.0 - \(V\) battery. After a long time elapses, the current in the coil is no longer changing. (a) What is the current in the coil? (b) What is the energy stored in the coil? (c) What is the rate of energy dissipation in the coil? (d) What is the induced emf in the coil?

Short Answer

Expert verified
Answer: The current in the coil is (2/11) A, the energy stored in the coil is (3/121) J, the rate of energy dissipation is (4/11) W, and the induced emf is 0 V.

Step by step solution

01

(a) Find the current in the coil

After a long time, the current in the coil no longer changes, and the system reaches a steady-state. In this case, the induced emf becomes zero, and the coil behaves like a regular resistor. Therefore, we can use Ohm's law to find the current in the coil: \(I = \dfrac{V}{R} = \dfrac{6 \, \text{V}}{33 \, \Omega} = \dfrac{2}{11} \, \text{A}\) So, the current in the coil is \(\dfrac{2}{11} \, \text{A}\).
02

(b) Calculate the energy stored in the coil

Using the given inductance and the calculated current, we can find the energy stored in the coil as follows: \(U = \dfrac{1}{2}LI^2 = \dfrac{1}{2}(0.15 \, \text{H})\left(\dfrac{2}{11} \, \text{A}\right)^2 = \dfrac{3}{121} \, \text{J}\) So, the energy stored in the coil is \(\dfrac{3}{121} \, \text{J}\).
03

(c) Find the rate of energy dissipation in the coil

To find the rate of energy dissipation, we use the following equation: \(P = I^2R = \left(\dfrac{2}{11} \, \text{A}\right)^2(33 \, \Omega) = \dfrac{4}{11} \, \text{W}\) So, the rate of energy dissipation in the coil is \(\dfrac{4}{11} \, \text{W}\).
04

(d) Calculate the induced emf in the coil

Since the coil has reached a steady-state and the current is no longer changing, there is no induced emf in the coil, which can be shown as: \(E = -L \dfrac{dI}{dt} = -0.15 \, \text{H} \times 0 = 0 \, \text{V}\) So, the induced emf in the coil is 0 V.

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