Chapter 29, Problem 49PQ

Three resistors with resistances R₁ = R/2 and R₂ = R₃ = R are connected as shown, and a potential difference of 225 V is applied across terminals a and b (Fig. P29.49).

a. If the resistor R₁ dissipates 75.0 W of power, what is the value of R?

b. What is the total power supplied to the circuit by the emf?

c. What is the potential difference across each of the three resistors?

To determine

Find the value of resistance from the given circuit.

Answer to Problem 49PQ

The value of $R$ is $\underset{\_}{338\Omega }$.

Explanation of Solution

Refer the figure $\text{P}29.49$; the equivalent resistance of the circuit is given by the parallel connection of $R_2$ and $R_3$ further connected in series to $R_1$.

Write the equivalent resistance of the parallel circuit as.

  $\frac{1}{R^{\text{'}}}=\frac{1}{R_2}+\frac{1}{R_3}$                                                                                                  (I)

Here $R^{\text{'}}$ is equivalent parallel resistance, $R_2$ is the resistance of second element and $R_3$ is the resistance of third element.

Since $R_1$ is parallel to the $R^{\text{'}}$.

Write the equivalent resistance of circuit as.

  $R_{eq}=R_1+R^{\text{'}}$                                                                                                 (II)

Here, $R_{eq}$ is the equivalent resistance of circuit.

Write the expression for the current drawn by the circuit from the voltage applied as.

  $I=\frac{\epsilon }{R_{eq}}$                                                                                                       (III)

Here $I$ is the current drawn and $\epsilon$ is the voltage supplied.

Write the expression for power dissipated across $R_1$ as

  $P=I^2{R}_1$                                                                                                  (IV)

Here, $P$ is the power dissipated across $R_1$.

Conclusion:

Substitute $R$ for $R_2$ and $R$ for $R_3$ in equation (I).

  $\begin{array}{c}\frac{1}{R^{\text{'}}}=\frac{1}{R}+\frac{1}{R}\\ =\frac{2}{R}\end{array}$

Rearrange the terms in above equation

  $R^{\text{'}}=\frac{R}{2}$

Substitute $\frac{R}{2}$ for $R^{\text{'}}$ and $\frac{R}{2}$ for $R_1$ in equation (II).

  $\begin{array}{c}{R}_{eq}=\frac{R}{2}+\frac{R}{2}\\ =R\end{array}$

Substitute $225\text{V}$ for $\epsilon$ and $R$ for $R_{eq}$ in equation (III).

  $I=\frac{225\text{V}}{R}$

Substitute $75\text{W}$ for $P$, $\frac{225\text{V}}{R}$ for $I$ and $\frac{R}{2}$ for $R_1$ in equation (IV).

  $\begin{array}{c}75\text{W}={\left(\frac{225\text{V}}{R}\right)}^2\times \frac{R}{2}\\ 75\text{W}=\frac{225\text{V}\times 225\text{V}}{2\times R}\end{array}$

Rearrange the terms in above equation

  $\begin{array}{c}R=\frac{225\text{V}\times 225\text{V}}{2\times 75\text{W}}\\ =337.5\Omega \\ \approx 338\Omega \end{array}$

Thus, the value of $R$ is $\underset{\_}{338\Omega }$.

To determine

Find the total power supplied to the circuit.

Answer to Problem 49PQ

The total power supplied to the circuit is $\underset{\_}{1.50\times {10}^2\text{W}}$.

Explanation of Solution

The total power supplied to the circuit is given as

  $P^{\text{'}}=\epsilon \times I$                                                                                                    (V)

Here $P^{\text{'}}$ is the power supplied to circuit and $I$ is the current supplied to circuit.

Conclusion:

Substitute $338\Omega$ for $R$ and $225\text{V}$ for $\epsilon$ in equation (III).

  $\begin{array}{c}I=\frac{225\text{V}}{338\Omega }\\ =0.666\text{A}\end{array}$

Substitute $0.666\text{A}$ for $I$ and $225\text{V}$ for $\epsilon$ in equation (V).

  $\begin{array}{c}{P}^{\text{'}}=225\text{V}\times \text{0}\text{.666}\text{A}\\ \text{=149}\text{.85}\text{W}\\ \text{=1}\text{.50}\times {\text{10}}^2\text{W}\end{array}$

Thus, the total power supplied to the circuit is $\underset{\_}{1.50\times {10}^2\text{W}}$.

To determine

Find the potential difference across the three resistors.

Answer to Problem 49PQ

The potential difference across the resistance $R_1$ is $\underset{\_}{113\text{V}}$, the potential difference across the resistance $R_2$ is $\underset{\_}{113\text{V}}$ and the potential difference across the resistance $R_3$ is $\underset{\_}{113\text{V}}$.

Explanation of Solution

Write the expression for the potential difference across the resistance $R_1$ as

  ${\epsilon }_1=I\times R_1$                                                                                                   (VI)

Here ${\epsilon }_1$ is the potential difference across resistance $R_1$.

Write the expression for the remaining potential drop is across the parallel combination of $R_2$ and $R_3$ as.

  ${\epsilon }^{\text{'}}=\epsilon -{\epsilon }_1$                                                                                                 (VII)

Here ${\epsilon }^{\text{'}}$ is the potential drop across the resistance $R_2$ and $R_3$.

Conclusion:

Substitute $0.666\text{A}$ for $I$ and $119\Omega$ for $R_1$ in the equation (VI)

  $\begin{array}{c}{\epsilon }_1=0.666\text{A}\times \text{119}\Omega \\ \text{=112}\text{.55}\text{V}\\ \approx \text{113}\text{V}\end{array}$

Substitute $225\text{V}$ for $\epsilon$ and $112.55\text{V}$ for ${\epsilon }_1$ in the equation (VII).

  $\begin{array}{c}{\epsilon }^{\text{'}}=225\text{V}-112.5\text{V}\\ \text{=112}\text{.5}\text{V}\\ \approx 113\text{V}\end{array}$

Thus, the potential difference across the resistance $R_1$ is $\underset{\_}{113\text{V}}$, the potential difference across the resistance $R_2$ is $\underset{\_}{113\text{V}}$ and the potential difference across the resistance $R_3$ is $\underset{\_}{113\text{V}}$.