Chapter 20, Problem 86E

(a)

To determine

The ratio of the power dissipated in the internal resistance to that of the load resistance.

Answer to Problem 86E

The ratio of the power dissipation in the internal resistance to the power dissipation in load is $1$.

Explanation of Solution

Write the expression for the power dissipation in the internal resistance.

    $P_r=\frac{{\epsilon }^{2}r}{{\left(R+r\right)}^2}$        (1)

Here, $P_r$ power across the internal resistance, $\epsilon$ is the battery EMF, $R$ is the load resistance and $r$ is the battery resistance.

Write the expression for the power dissipation across in the load.

    $P_R=\frac{{\epsilon }^{2}R}{{\left(R+r\right)}^2}$        (2)

Here, $P_R$ power across the load resistance.

Write the expression for the raio of the dissipated in the internal resistance to that of the load resistance.

    $P=\frac{P_r}{P_R}$        (3)

Conclusion:

Substitute $12\text{V}$ for $\epsilon$, $0.005\text{ohm}$ for $r$ and $0.005\text{ohm}$ for $R$ in equation (1).

    $\begin{array}{c}{P}_r=\frac{{\left(12\text{V}\right)}^2\left(0.005\text{ohm}\right)}{{\left(\left(0.005\text{ohm}\right)+\left(0.005\text{ohm}\right)\right)}^2}\\ =7200\text{W}\end{array}$

Substitute $12\text{V}$ for $\epsilon$, $0.005\text{ohm}$ for $r$ and $0.005\text{ohm}$ for $R$ in equation (2).

    $\begin{array}{c}{P}_R=\frac{{\left(12\text{V}\right)}^2\left(0.005\text{ohm}\right)}{{\left(\left(0.005\text{ohm}\right)+\left(0.005\text{ohm}\right)\right)}^2}\\ =7200\text{W}\end{array}$

Substitute $7200\text{W}$ for $P_r$ and $7200\text{W}$ for $P_R$ in equation (3).

    $\begin{array}{c}P=\frac{7200\text{W}}{7200\text{W}}\\ =1\end{array}$

Thus, the ratio of the power dissipation in the internal resistance to the power dissipation in load is $1$.

(b)

To determine

The ratio of the power dissipated in the internal resistance to that of the load resistance.

Answer to Problem 86E

The ratio of the power dissipation in the internal resistance to the power dissipation in load is $0.01$.

Explanation of Solution

Write the expression for the power dissipation in the internal resistance.

    $P_r=\frac{{\epsilon }^{2}r}{{\left(R+r\right)}^2}$

Write the expression for the power dissipation across in the load.

    $P_R=\frac{{\epsilon }^{2}R}{{\left(R+r\right)}^2}$

Write the expression for the raio of the dissipated in the internal resistance to that of the load resistance.

    $P=\frac{P_r}{P_R}$

Conclusion:

Substitute $12\text{V}$ for $\epsilon$, $0.005\text{ohm}$ for $r$ and $0.05\text{ohm}$ for $R$ in equation (1).

    $\begin{array}{c}{P}_r=\frac{{\left(12\text{V}\right)}^2\left(0.005\text{ohm}\right)}{{\left(\left(0.05\text{ohm}\right)+\left(0.005\text{ohm}\right)\right)}^2}\\ =238.01\text{W}\end{array}$

Substitute $12\text{V}$ for $\epsilon$, $0.005\text{ohm}$ for $r$ and $0.05\text{ohm}$ for $R$ in equation (2).

    $\begin{array}{c}{P}_R=\frac{{\left(12\text{V}\right)}^2\left(0.05\text{ohm}\right)}{{\left(\left(0.05\text{ohm}\right)+\left(0.005\text{ohm}\right)\right)}^2}\\ =2380.16\text{W}\end{array}$

Substitute $238.01\text{W}$ for $P_r$ and $2380.16\text{W}$ for $P_R$ in equation (3).

    $\begin{array}{c}P=\frac{238.01\text{W}}{2380.16\text{W}}\\ \approx 0.01\end{array}$

Thus, the ratio of the power dissipation in the internal resistance to the power dissipation in load is $0.01$.