The windings of an electromagnet have inductance \(L=8.0 \mathrm{H}\) and resistance \(R=2.0 \Omega . \quad A \quad 100.0-V \quad d c\) power supply is connected to the windings by closing switch \(S_{2} .\) (a) What is the current in the windings? (b) The electromagnet is to be shut off. Before disconnecting the power supply by opening switch \(S_{2},\) a shunt resistor with resistance \(20.0 \Omega\) is connected in parallel across the windings. Why is the shunt resistor needed? Why must it be connected before the power supply is disconnected? (c) What is the maximum power dissipated in the shunt resistor? The shunt resistor must be chosen so that it can handle at least this much power without damage. (d) When the power supply is disconnected by opening switch \(S_{2},\) how long does it take for the current in the windings to drop to \(0.10 \mathrm{A} ?\) (e) Would a larger shunt resistor dissipate the energy stored in the electromagnet faster? Explain.

Using Ohm's law (\(V = IR\)), we can find the current in the windings when switch \(S_2\) is closed: \(I = \frac{V}{R} = \frac{100.0 \,\mathrm{V}}{2.0 \,\Omega} = 50.0 \,\mathrm{A}\) So, the current in the windings is \(50.0 \, \mathrm{A}\).

When the electromagnet is turned off, the inductor will still have energy stored in it. The shunt resistor is needed to provide a path for the current to flow and dissipate the energy stored in the inductor as heat. It must be connected before the power supply is disconnected to avoid a sudden change in current, which could produce high voltage and possibly damage the circuit.

When the shunt resistor is first connected, the current will be maximum, and the power dissipated in the shunt resistor can be calculated using: \(P = I^2 R_s\) where \(R_s\) is the resistance of the shunt resistor. At this time, the total resistance in the circuit will be the parallel combination of \(R\) and \(R_s\): \(R_{total} = \frac{R \times R_s}{R + R_s} = \frac{2.0 \,\Omega \times 20.0 \,\Omega}{2.0 \,\Omega + 20.0 \,\Omega} = 1.818 \,\Omega\) Now we can find the current at this moment: \(I = \frac{V}{R_{total}} = \frac{100.0 \,\mathrm{V}}{1.818 \,\Omega} = 54.95 \,\mathrm{A}\) Now, let's determine how much current flows through the shunt resistor when it's first connected. Applying current divider rule: \(I_s = I \cdot \frac{R}{R + R_s} = 54.95 \, \mathrm{A} \cdot \frac{2.0 \,\Omega}{2.0 \,\Omega + 20.0 \,\Omega} = 4.995 \,\mathrm{A}\) Now, we can calculate the maximum power dissipated in the shunt resistor: \(P = I_s^2 R_s = (4.995 \,\mathrm{A})^2 \cdot 20.0 \,\Omega = 499.5 \,\mathrm{W}\) So the maximum power dissipation in the shunt resistor is \(499.5 \, \mathrm{W}\).

When the power supply is disconnected, the current in the windings will decay according to: \(I(t) = I_0 e^{-\frac{Rt}{L}}\) where \(I_0\) is the initial current, t is the time, and L is the inductance. We want to know the time it takes for the current to drop to 0.10 A: \(0.10 \,\mathrm{A} = 50.0 \,\mathrm{A} \cdot e^{-\frac{2.0 \,\Omega \cdot t}{8.0 \, H}}\) Solving for t: \(t = \frac{-8.0 \, H \cdot \ln{\frac{0.10 \,\mathrm{A}}{50.0 \,\mathrm{A}}}}{2.0 \,\Omega} = 18.4 \,\mathrm{s}\) So it takes \(18.4 \,\mathrm{s}\) for the current in the windings to drop to \(0.10 \,\mathrm{A}\).

A larger shunt resistor would result in less current flowing through it, therefore dissipating less power in the shunt resistor. This would lead to a slower energy dissipation rate from the electromagnet, and the whole energy stored will take longer to be dissipated. So, a larger shunt resistor would not dissipate the energy stored in the electromagnet faster.