Chapter 23, Problem 50P

(a)

To determine

The rate of deliver of energy

Answer to Problem 50P

The rate of deliver of energy is $66.0\text{W}$.

Explanation of Solution

The rate of deliver of energy is the power of the battery. Write the equation for the power of the battery.

    $P=I\Delta V$        (I)

Here, $I$ is the current and $\Delta V$ is the voltage of the battery.

Conclusion:

Substitute $3.00\text{A}$ for $I$ and $22.0\text{V}$ for $\Delta V$.

    $\begin{array}{c}P=\left(3.00\text{A}\right)\left(22.0\text{V}\right)\\ =66.0\text{W}\end{array}$        (II)

Therefore, the rate of deliver of energy by the battery is $66.0\text{W}$.

(b)

To determine

The power delivered to the resistance of the coil

Answer to Problem 50P

The power delivered to the resistance of the coil is $45.0\text{W}$.

Explanation of Solution

Write the equation for the power delivered to the resistance of the coil.

    $P=I\Delta V_R$        (III)

Here, $I$ is the current and $\Delta V_R$ is the voltage across the resistance.

Write the equation for the voltage across the resistance.

    $\Delta V_R=IR$        (IV)

Here, $I$ is the current and $R$ is the resistance of the resistor. Substitute equation (IV) in equation (III).

    $P=I^{2}R$

Conclusion:

Substitute $3.00\text{A}$ for $I$ and $5.00\Omega$ for $R$.

    $\begin{array}{c}P={\left(3.00\text{A}\right)}^2\left(5.00\Omega \right)\\ =45.0\text{W}\end{array}$        (V)

Therefore, the power delivered to the resistance of the coil is $45.0\text{W}$.

(c)

To determine

The rate of energy storage

Answer to Problem 50P

The rate of storage of energy is $21.0\text{W}$.

Explanation of Solution

Consider the inductor being ideal and connect in series with an ideal resistor. According to Kirchhoff’s voltage rule, the algebraic sum of all the voltages in any closed loop in a circuit is zero.

Write the equation for the algebraic sum of the voltages across the coil.

$+22.0\text{V}-\left(3.00\text{A}\right)\left(5.00\Omega \right)-\Delta V_L=0$        (VI)

Here, $+22.0\text{V}$ is the voltage across the battery, $\left(3.00\text{A}\right)\left(5.00\Omega \right)$ is the voltage across the resistor and $\Delta V_L$ is the voltage across the inductor.

The rate of storage of energy is the power. Write the equation for the power stored in the inductor.

$P=I\Delta V_L$        (VII)

Here, $I$ is the current in the current and $\Delta V_L$ is the voltage across the inductor.

Conclusion:

Rearrange equation (VI) and solve for $\Delta V_L$.

$\begin{array}{c}\Delta V_L=+22.0\text{V}-\left(3.00\text{A}\right)\left(5.00\Omega \right)\\ =7.00\text{V}\end{array}$

Substitute $3.00\text{A}$ for $I$ and $7.0\text{V}$ for $\Delta V_L$ in equation (VII).

$\begin{array}{c}P=\left(3.00\text{A}\right)\left(7.0\text{V}\right)\\ =21.0\text{W}\end{array}$        (VIII)

Therefore, the rate of storage of energy in the magnetic field is $21.0\text{W}$.

(d)

To determine

The relation between the three power values

Answer to Problem 50P

The power from the battery is the sum of the power across the internal resistance and the power in the magnetic field.

Explanation of Solution

From equation (II), the battery is delivering energy at a rate of $66.0\text{W}$.

From equation (V), the power delivered to the resistance of the coil is $45.0\text{W}$.

From equation (VIII), the rate of storage of energy in the magnetic field is $21.0\text{W}$.

From the value of different powers given in equation (II), equation (V) and equation (VIII), it can be inferred that $45.0\text{W}\text{+}21.0\text{W}\text{=}66.0\text{W}$

Conclusion:

Therefore, the power delivered from the battery is the sum of the power delivered to the internal resistance and the power stored in the magnetic field.

(e)

To determine

The validity of the relation

Answer to Problem 50P

Yes, it is valid in other instants as well

Explanation of Solution

The relation between the powers is that the power from the battery is the sum of the power across the internal resistance and the power in the magnetic field.

At any instant, the power generated by the battery is the sum of the power delivered to the internal resistance and the power stored in the magnetic field.

Conclusion:

Therefore, it is true that the relation between the power is valid at any istant.

(f)

To determine

The relation between the power at given instants

Answer to Problem 50P

The power delivered to the resistance is zero at $t=0$ and all the power is delivered to the resistance after several seconds.

Explanation of Solution

From equation (III) and equation (IV), write the equation for the power delivered to the resistance.

$P=I^{2}R$

Here, $I$ is the current and $R$ is the resistance of the resistor.

Write the equation for the power delivered by the magnetic field.

$P=L\frac{dI}{dt}$

Here, $L$ is the inductance of the coil and $I$ is the current which changes with time $t$.

Conclusion:

Immediately after $t=0$, the power $I^{2}R$ delivered to the internal resistance of the coil is almost zero. At this moment the power delivered to the magnetic field $L\left(dI/dt\right)$ is high that almost all the power generated from the battery is stored in the magnetic field.

After some time, the current does not change anymore and hence there is no power being stored in the magnetic field. All the power from the battery is delivered to the resistance of the coil.