Chapter 29, Problem 39PQ

(a)

To determine

Find the current through the emf device and each resistor in circuit 1.

Answer to Problem 39PQ

The current through the Emf device and each resistor in circuit 1 is $\underset{\_}{0.917\text{A}}$. The current for the given circuit resistance is $\underset{\_}{0.500\text{A}}$ for $R_1$, $\underset{\_}{0.250\text{A}}$ for $R_2$ and is $\underset{\_}{0.167\text{A}}$ for $R_3$.

Explanation of Solution

According to Kirchhoff’s junction rule, in any junction, the sum of the all the currents entering the junction equals the sum of all the currents exiting the junction.

Redraw the circuit 1 and labeled it as given below

In parallel circuit, voltage across all three resistors is same that means potential difference between node A and node B is same ($V_A-V_B=\epsilon$). Current will be divide equally in all three branches.

According to Ohm’s law,

  $\epsilon =I{R}_{\text{eq}}$                                                                                                          (I)

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

Rearrange the equation (I) in terms of total current

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

Write the expression for equivalent resistance as.

    $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Rearrange the above expression.

    $R_{eq}=\left(\frac{R_1{R}_2{R}_3}{R_1{R}_2+R_2{R}_3+R_1{R}_3}\right)$                                                                         (III)

Write the expression for current $I_1$.

    $I_1=\frac{\epsilon }{R_1}$                                                                                                      (IV)

Here, $I_1$ is current in first branch, and $R_1$ is the resistance of first branch.

Write the expression for current $I_2$

  $I_2=\frac{\epsilon }{R_2}$                                                                                                    (V)

Here, $I_2$ is current in second branch, and $R_2$ is the resistance of second branch.

Write the expression for current $I_3$

  $I_3=\frac{\epsilon }{R_3}$                                                                                                     (VI)

Here, $I_3$ is current in third branch, and $R_3$ is the resistance of third branch.

Conclusion:

Substitute $R\Omega$ for $R_1$, $2R\Omega$ for $R_2$ and $3R\Omega$ for $R_3$ in equation (III)

    $\begin{array}{c}{R}_{\text{eq}}=\left(\frac{R\left(2R\right)\left(3R\right)}{R\left(2R\right)+R\left(3R\right)+2R\left(3R\right)}\right)\\ =\left(\frac{6R^3}{11R^2}\right)\\ =\left(\frac{6R}{11}\right)\end{array}$

Substitute $15\Omega$ for $R$. in above equation

    $\begin{array}{c}{R}_{eq}=\left(\frac{6\left(15\Omega \right)}{11}\right)\\ =8.18\Omega \end{array}$

Substitute $8.18\Omega$ for $R_{eq}$ and $7.50\text{ V}$ for $\epsilon$ in equation (II).

    $\begin{array}{c}I=\frac{7.50\text{ V}}{8.18\Omega }\\ =0.9168\text{ A}\\ \approx 0.9167\text{ A}\end{array}$

Thus, the current in circuit 1 is $\underset{\_}{0.9167\text{ A}}$.

Substitute $7.50\text{ V}$ for $\epsilon$ and $15\Omega$ for $R_1$ in equation (IV).

    $\begin{array}{c}{I}_1=\frac{7.50\text{ V}}{15.0\Omega }\\ =0.500\text{ A}\end{array}$

Substitute $7.50\text{ V}$ for $\epsilon$ and $30\Omega$ for $R_2$ in equation (V).

    $\begin{array}{c}{I}_2=\frac{7.50\text{ V}}{30.0\Omega }\\ =0.250\text{ A}\end{array}$

Substitute $7.50\text{ V}$ for $\epsilon$ and $\text{45 }\Omega$ for $R_3$ in equation (VI).

    $\begin{array}{c}{I}_3=\frac{7.50\text{ V}}{45.0\Omega }\\ =0.167\text{ A}\end{array}$

Thus, the current through the Emf device and each resistor in circuit 1 is $\underset{\_}{0.9167\text{ A}}$. The current for the given circuit resistance is $\underset{\_}{0.500\text{ A}}$ for $R_1$, $\underset{\_}{0.250\text{ A}}$ for $R_2$ and is $\underset{\_}{0.167\text{ A}}$ for $R_3$.

(b)

To determine

Find the current through the emf device and each resistor in circuit 2 refer to figure P29.28.

Answer to Problem 39PQ

The current through the Emf device and each resistor in circuit 2 is $\underset{\_}{0.9167\text{ A}}$. The current for the given circuit resistance is $\underset{\_}{0.500\text{ A}}$ for $R_1$, $\underset{\_}{0.250\text{ A}}$ for $R_2$ and is $\underset{\_}{0.167\text{ A}}$ for $R_3$.

Explanation of Solution

According to Kirchhoff’s junction rule, in any junction, the sum of the all the currents entering the junction equals the sum of all the currents exiting the junction.

Redraw the circuit 2 and labeled it as given below.

In parallel circuit 2, voltage across all three resistors is same that means potential difference between node a and node b, node c & node d and node e & node f are same $\left(V_{AB}=V_{CD}=V_{EF}=\epsilon \right)$. Current will be divide equally in all three branches.

Write the expression for current $I_1$ as.

    $I_1=\frac{V_{AB}}{R_1}$                                                                                            (VII)

Here, $I_1$ is the current in branch 1 , $V_{AB}$ is the voltage drop across $ab$ and $R_1$ is the resistance.

Write the expression for current $I_2$ as.

  $I_2=\frac{V_{CD}}{R_2}$                                                                                            (VIII)

Here, $I_2$ is the current in branch 2 , $V_{CD}$ is the voltage drop across $cd$ and $R_2$ is the resistance.

Write the expression for current $I_3$ as.

  $I_3=\frac{V_{EF}}{R_3}$                                                                                              (IX)

Here, $I_3$ is the current in branch 3 , $V_{EF}$ is the voltage drop across $ef$ and $R_3$ is the resistance.

Conclusion:

Substitute $R\Omega$ for $R_1$, $2R\Omega$ for $R_2$ and $3R\Omega$ for $R_3$ in equation in equation (III)

    $\begin{array}{c}\frac{1}{R_{eq}}=\left(\frac{2R\left(3R\right)+R\left(3R\right)+R\left(2R\right)}{R\left(2R\right)\left(3R\right)}\right)\\ \text{=}\left(\frac{6R^2+3R^2+2R^2}{6R^3}\right)\\ =\left(\frac{11R^2}{6R^3}\right)\\ =\left(\frac{11}{6R}\right)\end{array}$

Substitute $15\Omega$ for $R$. in above equation

    $\begin{array}{c}\frac{1}{R_{eq}}=\frac{11}{6\left(15\Omega \right)}\\ =\frac{11}{90\Omega }\\ R_{eq}=8.18\Omega \end{array}$

Substitute $8.18\Omega$ for $R_{eq}$ and $7.50\text{ V}$ for $\epsilon$ in equation (II)

    $\begin{array}{c}I=\frac{7.50\text{ V}}{8.18\Omega }\\ =0.9168\\ \approx 0.9167\text{ A}\end{array}$

Thus, the current in circuit 1 is $\underset{\_}{0.9167\text{ A}}$.

Substitute $7.50\text{ V}$ for $V_{AB}$ and $15\Omega$ for $R_1$ in equation (VII).

    $\begin{array}{c}{I}_1=\frac{7.50\text{ V}}{15.0\Omega }\\ =0.500\text{ A}\end{array}$

Substitute $7.50\text{ V}$ for $V_{CD}$ and $\text{30 }\Omega$ for $R_2$ in equation (VIII).

    $\begin{array}{c}{I}_2=\frac{7.50\text{ V}}{30.0\Omega }\\ =0.250\text{ A}\end{array}$

Substitute $7.50\text{ V}$ for $V_{EF}$ and $45\Omega$ for $R_3$ in equation (IX)

    $\begin{array}{c}{I}_3=\frac{7.50\text{ V}}{45.0\Omega }\\ =0.167\text{ A}\end{array}$

Thus, the current for the given circuit resistance is $\underset{\_}{0.500\text{ A}}$ for $R_1$, $\underset{\_}{0.250\text{ A}}$ for $R_2$ and is $\underset{\_}{0.167\text{ A}}$ for $R_3$.