Chapter 20, Problem 60P

(a)

To determine

The current in the windings.

Answer to Problem 60P

The current in the windings is $\underset{\_}{50\text{A}}$.

Explanation of Solution

Given that the voltage offered by the power supply is $100.0\text{V}$, and he resistance of the winding is $2.0\Omega$.

The equilibrium current in the circuit is equal to the maximum current flowing through the winding.

Write the expression for the maximum current through the winding.

$I_0=\frac{{\epsilon }_{\text{b}}}{R}$ (I)

Here, $I_0$ is the maximum current, ${\epsilon }_{\text{b}}$ is the emf offered by the battery, and $R$ is the resistance of the winding.

Conclusion:

Substitute $100.0\text{V}$ for ${\epsilon }_{\text{b}}$, and $2.0\Omega$ for $R$ in equation (I) to find $I_{\text{0}}$.

$\begin{array}{c}{I}_0=\frac{100.0\text{V}}{2.0\Omega }\\ =50\text{A}\end{array}$

Therefore, the current in the windings is $\underset{\_}{50\text{A}}$.

(b)

To determine

The need of the shunt resistor in the electromagnet and the reason for which it is connected before disconnecting the power supply.

Answer to Problem 60P

The shunt resistor reduces the rate of change of current in the winding of the electromagnet, and hence allowing the electromagnet to shut off safely. The shunting action is possible only if the shunt resistor is connected in the circuit before disconnecting the power supply.

Explanation of Solution

If the power supply to the electromagnet is stopped suddenly, then the induced emf in the windings will be very large so that there is a possibility of damaging of the winding itself. Moreover, it is likely that sparks would complete the circuit across the open switch.

The presence of a shunt resistor reduces the rate of change of current in the winding of the electromagnet, and hence allowing the electromagnet to shut off safely. In order to have the effect of the shunt resistor in the process, it must be present in the circuit before switching off the power supply. Simply, the shunt resistor must be connected before disconnecting the power supply.

Conclusion:

Therefore, the shunt resistor reduces the rate of change of current in the winding of the electromagnet, and hence allowing the electromagnet to shut off safely. The shunting action is possible only if the shunt resistor is connected in the circuit before disconnecting the power supply.

(c)

To determine

The maximum power dissipated in the shunt resistor.

Answer to Problem 60P

The maximum power dissipated in the shunt resistor is $\underset{\_}{50\text{kW}}$.

Explanation of Solution

Given that the resistance of the shunt resistor is $20.0\Omega$. It is obtained that the maximum current in the circuit is $50\text{A}$.

Write the expression for the maximum power dissipated in a resistor.

$P_{\text{max}}=I_0^{2}R$ (II)

Here, $P_{\text{max}}$ is the maximum power dissipated.

Conclusion:

Substitute $50\text{A}$ for $I_0$, and $20.0\Omega$ for $R$ in equation (II) to find $P_{\text{max}}$.

$\begin{array}{c}{P}_{\text{max}}={\left(50\text{A}\right)}_0^2\left(20.0\Omega \right)\\ =50\times {10}^3\text{W}\\ =50\times {10}^3\text{W}\times \frac{1\text{kW}}{1000\text{W}}\\ =50\text{kW}\end{array}$

Therefore, the maximum power dissipated in the shunt resistor is $\underset{\_}{50\text{kW}}$.

(d)

To determine

The time taken for the current in the windings to drop to $0.10\text{A}$ after disconnecting the power supply.

Answer to Problem 60P

The time taken for the current in the windings to drop to $0.10\text{A}$ after disconnecting the power supply is $\underset{\_}{2.3\text{s}}$.

Explanation of Solution

Given that the resistance of the shunt resistor is  $20.0\Omega$, the resistance of the windings is $2.0\Omega$, the inductance is $8.0\text{H}$, the maximum current is $50\text{A}$, and the current at time $t$ is $0.10\text{A}$.

When the switch $S_2$ is opened, the circuit becomes an $RL$ circuit.

Write the expression for the current in an $RL$ circuit.

$I\left(t\right)=I_0{e}^{-t/\tau }$ (III)

Here, $I\left(t\right)$ is the current, $t$ is the time, and $\tau$ is the time constant of the circuit.

Solve equation (III) for $t$.

$\begin{array}{l}\ln \frac{I}{I_0}=-\frac{t}{\tau }\\ t=\tau \ln \frac{I_0}{I}\end{array}$ (IV)

Write the expression for the time constant of the $RL$ circuit.

$\tau =\frac{L}{R_{\text{eq}}}$ (V)

Here, $L$ is the inductance, $R_{\text{eq}}$ is the equivalent resistance.

Write the expression for the equivalent resistance in the given circuit.

$R_{\text{eq}}=R_{\text{winding}}+R_{\text{shunt}}$ (VI)

Use equation (VI) in (V).

$\tau =\frac{L}{R_{\text{winding}}+R_{\text{shunt}}}$ (VII)

Use equation (VII) in (IV).

$t=\left(\frac{L}{R_{\text{winding}}+R_{\text{shunt}}}\right)\ln \frac{I_0}{I}$ (VIII)

Conclusion:

Substitute $20.0\Omega$ for $R_{\text{shunt}}$, $2.0\Omega$ for $R_{\text{winding}}$, $8.0\text{H}$ for $L$, $50\text{A}$ for $I_0$, and $0.10\text{A}$ for $I$ in equation (VIII) to find $t$.

$\begin{array}{c}t=\left(\frac{8.0\text{H}}{2.0\Omega +20.0\Omega }\right)\ln \frac{50\text{A}}{0.10\text{A}}\\ =2.3\text{s}\end{array}$

Therefore, the time taken for the current in the windings to drop to $0.10\text{A}$ after disconnecting the power supply is $\underset{\_}{2.3\text{s}}$.

(e)

To determine

Whether a large shunt resistor would dissipate the energy stored in the electromagnet faster or not.

Answer to Problem 60P

A large shunt resistor would dissipate the energy stored in the electromagnet faster.

Explanation of Solution

According to equation (VIII), the time taken for the dissipation of current in the winding is inversely proportional to the equivalent resistance of the circuit.

$t\propto \frac{1}{R_{\text{winding}}+R_{\text{shunt}}}$

If the shunt resistance value is increased, then the equivalent resistance of the circuit increases. This results the time taken for the current dissipation in the winding or the electromagnet decrease. That is, a large shunt resistor would dissipate the energy stored in the electromagnet faster

Conclusion:

Therefore, a large shunt resistor would dissipate the energy stored in the electromagnet faster.