Chapter 19, Problem 17P

(a)

To determine

The equivalent resistance of the circuit.

Answer to Problem 17P

The equivalent resistance of the circuit is $841.73\Omega$ .

Explanation of Solution

Given info:

Resistance of the first resistor in parallel in circuit, $R_1=820\Omega$

Resistance of the second resistor in parallel in circuit, $R_2=680\Omega$

Resistance of the resistor in series in circuit, $R_3=470\Omega$

Formula Used:

The expression to calculate the equivalent resistance of the circuit is,

  $R_e=\frac{R_1{R}_2}{R_1+R_2}+R_3$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_e=\frac{\left(820\Omega \right)\left(680\Omega \right)}{820\Omega +680\Omega }+470\Omega \\ =841.73\Omega \end{array}$

Conclusion:

Therefore, the equivalent resistance of the circuit is $841.73\Omega$ .

(b)

To determine

The voltage across each resistor.

Answer to Problem 17P

The voltage across each resistor in parallel connection in the circuit is $5.30\text{V}$ and the voltage across series resistance is $6.70\text{V}$ .

Explanation of Solution

Given info:

The voltage of the battery in the circuit is, $V=12\text{V}$ .

Formula Used:

The expression to calculate the current in the circuit is,

  $I=\frac{V}{R_e}$

The expression to calculate the voltage across series resistance is,

  $V_3=I{R}_3$

The expression to calculate the voltage across each resistance of parallel resistors is,

  $V_{1/2}=I\left(\frac{R_1{R}_2}{R_1+R_2}\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}I=\frac{12\text{V}}{841.73\Omega }\\ =0.0143\text{A}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{V}_3=\left(0.0143\text{A}\right)\left(470\Omega \right)\\ =6.70\text{V}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{V}_{1/2}=\left(0.0143\text{A}\right)\left[\frac{\left(820\Omega \right)\left(680\Omega \right)}{820\Omega +680\Omega }\right]\\ =5.30\text{V}\end{array}$

Conclusion:

Thus, the voltage across each resistor in parallel connection in the circuit is $5.30\text{V}$ and the voltage across series resistance is $6.70\text{V}$ .