Chapter 29, Problem 43PQ

The emfs in Figure P29.43 are ε₁ = 6.00 V and ε₂ = 12.0 V. The resistances are R₁ = 15.0 Ω, R₂ = 30.0 Ω, R₃ = 45.0 Ω, and R₄ = 60.0 Ω. Find the current in each resistor when the switch is

1. a. open and
2. b. closed.

To determine

The current in each resistor when the switch is open.

Answer to Problem 43PQ

The current flowing through the resistor $R_1$ is $0.133\text{A}$, $R_2$ is $0.133\text{A}$, $R_3$ is $0.133\text{A}$ and $R_4$ is $0\text{A}$.

Explanation of Solution

As the switch is kept open; the current in the wire EFAB will be zero as the circuit is disconnected.

The resistors in the loop BCDEB are connected in series. The current flowing in all the elements in a series circuit is constant.

Write the expression for the equivalent resistance in the loop BCDEB as.

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

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

Write the expression for the current in the loop BCDEB as.

    $I=\frac{{\epsilon }_2}{R_{eq}}$                                                                                                                    (II)

Here, $I$ is the current in the loop BCDEB and ${\epsilon }_2$ is the Emf of the battery.

No current flows through the resistor $R_4$ as the switch is kept open and therefore the value of the current $I_0$ in the loop ABEFA is zero.

Conclusion:

Substitute $15\text{Ω}$ for $R_1$ , $30\text{Ω}$ for $R_2$ and $45\text{Ω}$ for $R_3$ in equation (I).

    $\begin{array}{c}{R}_{eq}=15\text{Ω}+30\text{Ω}+45\text{Ω}\\ =9\text{0}\text{Ω}\end{array}$

Substitute $12\text{V}$ for ${\epsilon }_2$ and $90\text{Ω}$ for $R_{eq}$ in equation (II).

  $\begin{array}{c}I=\frac{12\text{V}}{90\text{Ω}}\\ =0.133\text{A}\end{array}$

Thus, the current flowing through the resistor $R_1$ is $0.133\text{A}$, $R_2$ is $0.133\text{A}$, $R_3$ is $0.133\text{A}$ and $R_4$ is $0\text{A}$.

To determine

The current in each resistor when the switch is closed.

Answer to Problem 43PQ

The current flowing through the resistor $R_1$ is $0.179\text{A}$, $R_2$ is $0.124\text{A}$, $R_3$ is $0.124\text{A}$ and $R_4$ is $0.0552\text{A}$.

Explanation of Solution

The switch is closed, the current starts flowing through all the resistors in the circuit.

The circuit diagram for the closed switch is shown below.

Write the expression for the Kirchhoff’s voltage law in loop ABCDEFA.

    $-{\epsilon }_1-I^{\text{'}}{R}_3+{\epsilon }_2-I^{\text{'}}{R}_2+I_o{R}_4=0$                                                                               (III)

Here, $I^{\text{'}}$ is the current flowing in loop BCDEB and $I_o$ is the current flowing in loop ABEFA, $R_4$ is the resistance of fourth element and ${\epsilon }_1$ is the Emf of battery.

Write the expression for the Kirchhoff’s voltage law in loop ABEFA.

  $-{\epsilon }_1+\left(I_o+I^{\text{'}}\right)R_1+I_o{R}_4=0$                                                                                     (IV)

Write the expression for the current through resistor $R_1$.

  $I_{R_1}=I_o+I^{\text{'}}$                                                                                                               (V)

Here, $I_{R_1}$ is the current in resistor $R_1$.

Write the expression for the current through resistor $R_2$.

  $I_{R_2}=I^{\text{'}}$                                                                                                                    (VI)

Here, $I_{R_2}$ is the current in resistor $R_2$.

Write the expression for the current through resistor $R_3$.

  $I_{R_3}=I^{\text{'}}$                                                                                                                   (VII)

Here, $I_{R_3}$ is the current in resistor $R_3$.

Write the expression for the current through resistor $R_4$.

  $I_{R_4}=I_o$                                                                                                                 (VIII)

Here, $I_{R_4}$ is the current in resistor $R_4$.

Conclusion:

Substitute $6\text{V}$ for ${\epsilon }_1$ , $45\text{Ω}$ for $R_3$$12\text{V}$ for ${\epsilon }_2$ , $30\text{Ω}$ for $R_2$ and $60\text{Ω}$ for $R_4$ in equation (III).

  $\begin{array}{c}-\left(6\text{V}\right)-I^{\text{'}}\left(45\text{Ω}\right)+\left(12\text{V}\right)-I^{\text{'}}\left(30\text{Ω}\right)+I_o\left(60\text{Ω}\right)=0\\ 75I^{\text{'}}-60I_o=6\end{array}$

Rearrange the above equation for $I^{\text{'}}$ .

  $I^{\text{'}}=\frac{6+60I_o}{75}$                                                                                                           (IX)

Substitute $6\text{V}$ for ${\epsilon }_1$, $15\text{Ω}$ for $R_1$ and $60\text{Ω}$ for $R_4$ in equation (IV).

  $\begin{array}{c}-\left(6\text{V}\right)+\left(I_o+I^{\text{'}}\right)\left(15\text{Ω}\right)+I_o\left(60\text{Ω}\right)=0\\ 75I_o+15I^{\text{'}}=6\end{array}$

Substitute $\frac{6+60I_o}{75}$ for $I^{\text{'}}$ in above equation.

  $\begin{array}{c}75I_o+15\left(\frac{6+60I_o}{75}\right)=6\\ 75I_o+12I^{\text{'}}=\frac{24}{5}\end{array}$

Rearrange the above expression for $I_o$.

    $I_o=\text{0}\text{.0552}\text{A}$

Substitute $0.0552\text{A}$ for $I_o$ in equation (IX).

  $\begin{array}{c}{I}^{\text{'}}=\frac{6+\left(\left(60\right)\left(0.0552\text{A}\right)\right)}{75}\\ =0.124\text{A}\end{array}$

Substitute $\text{0}\text{.0552}\text{A}$ for $I_o$ and $0.124\text{A}$ for $I^{\text{'}}$ in equation (V).

  $\begin{array}{c}{I}_{R_1}=\left(\text{0}\text{.0552}\text{A}\right)+\left(0.124\text{A}\right)\\ =0.179\text{A}\end{array}$

Substitute $0.124\text{A}$ for $I^{\text{'}}$ in equation (VI).

  $I_{R_2}=0.124\text{A}$

Substitute $0.124\text{A}$ for $I^{\text{'}}$ in equation (VII).

  $I_{R_3}=0.124\text{A}$

Substitute $\text{0}\text{.0552}\text{A}$ for $I_o$ in equation (V).

  $I_{R_4}=\text{0}\text{.0552}\text{A}$

Thus, the current flowing through the resistor $R_1$ is $0.179\text{A}$, $R_2$ is $0.124\text{A}$, $R_3$ is $0.124\text{A}$ and $R_4$ is $0.0552\text{A}$.