Chapter 29, Problem 15PQ

(a)

To determine

The current through the each resistor.

Answer to Problem 15PQ

The current through the each resistor is $\underset{\_}{0.339\text{ A}}$.

Explanation of Solution

Refer to Fig 29.15; in this series circuit, current is same for entire closed loop.

Write the expression for the current in the circuit as.

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

Here, $I$ is the circuit current, $\epsilon$ is the emf of the circuit and $R_{eq}$ is the equivalent resistance of the circuit.

Write the expression for equivalent resistance of three resistors connected in series as.

  $R_{eq}=R_1+R_2+R_3$                                                                                        (II)

Here, $R_{eq}$ is the equivalent resistance of the circuit, $R_1$ is the resistance of first element, $R_2$ is the resistance of first element and $R_3$ is the resistance of first element.

Conclusion:

Substitute $13.4\Omega$ for $R_1$ , $20.5\Omega$ for $R_2$ and $9.8\Omega$ for $R_3$ in equation (II).

    $\begin{array}{c}{R}_{eq}=13.4\Omega +20.5\Omega +9.8\Omega \\ =43.7\Omega \end{array}$

Substitute $43.7\Omega$ for $R_{eq}$ and $14.8\text{ V}$ for $\epsilon$ in equation (I).

    $\begin{array}{c}I=\frac{14.8\text{ V}}{43.7\Omega }\\ =0.339\text{ A}\end{array}$

Thus, the current passing through each resistor is $\underset{\_}{0.339\text{ A}}$

(b)

To determine

Find the voltage across each resistor

Answer to Problem 15PQ

The voltage across each resistor is $\underset{\_}{4.54\text{ V}}$ for $R_1$, $\underset{\_}{6.95\text{ V}}$ for $R_2$ and $\underset{\_}{3.32\text{ V}}$ for $R_3$.

Explanation of Solution

Write the expression for voltage across each resistance when resistors connected in series as.

    $\epsilon =IR$                                                                                                          (III)

Here, $R$ is the resistance of the circuit.

Conclusion:

Substitute $0.339\text{ A }$ for $I$ and $13.4\Omega$ for $R_1$ in equation (III).

    $\begin{array}{c}{\epsilon }_1=\left(0.339\text{ A}\right)\left(\text{13}\text{.4 }\Omega \right)\\ =\text{4}\text{.54 V}\end{array}$

Substitute $0.339\text{ A }$ for $I$ and $20.5\Omega$ for $R_2$ in equation (III).

    $\begin{array}{c}{\epsilon }_2=\left(0.339\text{ A}\right)\left(\text{20}\text{.5 }\Omega \right)\\ =6.\text{95 V}\end{array}$

Substitute $0.339\text{ A }$ for $I$ and $9.80\Omega$ for $R_3$ in equation (III).

    $\begin{array}{c}{\epsilon }_3=\left(0.339\text{ A}\right)\left(\text{9}\text{.80 }\Omega \right)\\ =\text{3}\text{.32 V}\end{array}$

Thus, the voltage across each resistor is $\underset{\_}{4.54\text{ V}}$ for $R_1$, $\underset{\_}{6.95\text{ V}}$ for $R_2$ and $\underset{\_}{3.32\text{ V}}$ for $R_3$.

(c)

To determine

Find the power is consumed by each resistor

Answer to Problem 15PQ

The power consumed by each resistor is $\underset{\_}{1.54\text{ W}}$ for $R_1$, $\underset{\_}{2.36\text{ W}}$ for $R_2$ and $\underset{\_}{1.13\text{ W}}$ for $R_3$.

Explanation of Solution

Write the expression for power consumed by resistor as

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

Here, $I$ is the circuit current, $P$ is the Power consumed by the element in the circuit and $R$ is the resistance of the circuit.

Write the expression for the power consumed by Emf device as.

    $P_d=\epsilon I$                                                                                                           (V)

Here, $P_d$ is the power consumed by device and $\epsilon$ is the emf of device.

Conclusion:

Substitute $0.339\text{ A }$ for $I$ and $13.4\Omega$ for $R_1$ in equation (IV)

    $\begin{array}{c}{P}_1={\left(0.339\text{ A}\right)}^2\left(13.4\Omega \right)\\ =1.54\text{ W}\end{array}$

Substitute $0.339\text{ A }$ for $I$ and $20.5\Omega$ for $R_2$ in equation (IV)

    $\begin{array}{c}{P}_2={\left(0.339\text{ A}\right)}^2\left(20.5\Omega \right)\\ =2.36\text{ W}\end{array}$

Substitute $0.339\text{ A }$ for $I$ and $9.80\Omega$ for $R_3$ in equation (IV)

    $\begin{array}{c}{P}_2={\left(0.339\text{ A}\right)}^2\left(9.80\Omega \right)\\ =1.13\text{ W}\end{array}$

Substitute $0.339\text{ A}$ for $I$ and $14.8\text{ V}$ for $\epsilon$ in equation (V).

    $\begin{array}{c}{P}_d=\left(0.339\text{A}\right)\left(14.8\text{ V}\right)\\ =5.02\text{ W}\end{array}$

Thus, the power consumed by each resistor is $\underset{\_}{1.54\text{ W}}$ for $R_1$, $\underset{\_}{2.36\text{ W}}$ for $R_2$, $\underset{\_}{1.13\text{ W}}$ for $R_3$ and $5.02\text{ W}$ for emf device.

(d)

To determine

the current , voltage drop and power through each resistor when $R_3$ is replaced by twice of $2R_3$.

Answer to Problem 15PQ

The current through each resistor when $R_3$ is replaced by twice of $2R_3$. is $\underset{\_}{0.277\text{ A}}$. The power consumed by $R_1$ is $\underset{\_}{1.03\text{ W}}$, $R_2$ is $\underset{\_}{1.57\text{ W}}$ and $R_3$ is $\underset{\_}{1.50\text{ W}}$. The voltage drop across $R_1$ is $\underset{\_}{3.70\text{ V}}$, $R_2$ is  $\underset{\_}{5.67\text{ V}}$ and $R_3$ is $\underset{\_}{5.42\text{ V}}$.

Explanation of Solution

Conclusion:

Substitute $13.4\Omega$ for $R_1$ , $20.5\Omega$ for $R_2$ and $19.6\Omega$ for $R_3$ in equation (II).

    $\begin{array}{c}{R}_{eq}=13.4\Omega +20.5\Omega +19.6\Omega \\ =53.5\Omega \end{array}$

Substitute $\text{53}\text{.5 }\Omega$ for $R_{eq}$ and $14.8\text{ V}$ for $\epsilon$ in equation (I)

    $\begin{array}{c}I=\frac{14.8\text{ V}}{\text{53}\text{.5 }\Omega }\\ =0.2766\text{ A}\\ \approx \text{0}\text{.277}\text{A}\end{array}$

Thus, the current passing through each resistor is $\underset{\_}{0.277\text{ A}}$

Substitute $0.2766\text{ A }$ for $I$ and $13.4\Omega$ for $R_1$ in equation (IV)

    $\begin{array}{c}{P}_1={\left(0.2766\text{ A}\right)}^2\left(13.4\Omega \right)\\ =1.025\text{ W}\\ \approx 1.03\text{ W}\end{array}$

Substitute $0.2776\text{ A }$ for $I$ and $20.5\Omega$ for $R_2$ in equation (IV)

    $\begin{array}{c}{P}_2={\left(0.2766\text{ A}\right)}^2\left(20.5\Omega \right)\\ =1.568\text{ W}\\ \approx \text{1}\text{.57}\text{W}\end{array}$

Substitute $0.2766\text{ A }$ for $I$ and $19.6\Omega$ for $R_3$ in equation (IV)

    $\begin{array}{c}{P}_3={\left(0.2766\text{ A}\right)}^2\left(19.60\Omega \right)\\ =1.499\text{ W}\\ \approx 1.50\text{ W}\end{array}$

Substitute $0.2766\text{ A}$ for $I$ and $14.8\text{ V}$ for $\epsilon$ in equation (V).

    $\begin{array}{c}{P}_d=\left(0.2766\text{A}\right)\left(14.8\text{ V}\right)\\ =4.09\text{ W}\end{array}$

Thus, the power consumed by each resistor is $\underset{\_}{1.03\text{ W}}$ for $R_1$, $\underset{\_}{1.57\text{ W}}$ for $R_2$, $\underset{\_}{1.50\text{ W}}$ for $R_3$ and $4.09\text{ W}$ for emf device.

Substitute $0.2766\text{ A }$ for $I$ and $13.4\Omega$ for $R_1$ in equation (III).

    $\begin{array}{c}{\epsilon }_1=\left(0.2766\text{ A}\right)\left(\text{13}\text{.4 }\Omega \right)\\ =3.\text{71 V}\end{array}$

Substitute $0.2766\text{ A }$ for $I$ and $20.5\Omega$ for $R_2$ in equation (III).

    $\begin{array}{c}{\epsilon }_2=\left(0.2766\text{ A}\right)\left(\text{20}\text{.5 }\Omega \right)\\ =\text{5}\text{.67 V}\end{array}$

Substitute $0.2766\text{ A }$ for $I$ and $19.60\Omega$ for $R_3$ in equation (III).

    $\begin{array}{c}{\epsilon }_3=\left(0.2766\text{ A}\right)\left(1\text{9}\text{.60 }\Omega \right)\\ =5.\text{42 V}\end{array}$

Thus, the voltage drop across $R_1$ is $\underset{\_}{3.71\text{ V}}$, $R_2$ is  $\underset{\_}{5.67\text{ V}}$ and $R_3$ is $\underset{\_}{5.42\text{ V}}$.