Chapter 8, Problem 22E

(a)

To determine

The current passing through a parallel circuit.

Answer to Problem 22E

The current passing through a parallel circuit is $\text{10 A}$.

Explanation of Solution

Given info: First resistor having resistance $10\Omega$, second resistor having resistance $\text{15 }\Omega$, third resistor having resistance $\text{20 }\Omega$ and dc voltage source is $\text{50 V}$.

Explanation:

In a parallel circuit two resistors are arranged with their heads connected to head and tails are connected tails across the dc voltage source as shown in below figure.

Formula to calculate the equivalent resistance in a parallel circuit is,

$\begin{array}{l}\frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\ R_P=\frac{\left(R_1\right)\left(R_2\right)\left(R_3\right)}{R_2{R}_3+R_1{R}_3+R_1{R}_2}\end{array}$

Here,

$R_1,R_2,R_3$ are the resistances for individual resistors

Substitute $\text{10 }\Omega$ for $R_1$ and $\text{15 }\Omega$ for $R_2$ and $\text{20 }\Omega$ for $R_3$.

$\begin{array}{c}{R}_P=\frac{\left(\text{10 }\Omega \right)\left(\text{15 }\Omega \right)\left(\text{20 }\Omega \right)}{\left(\text{15 }\Omega \right)\left(\text{20 }\Omega \right)+\left(\text{10 }\Omega \right)\left(\text{20 }\Omega \right)+\left(\text{10 }\Omega \right)\left(\text{15 }\Omega \right)}\\ =\frac{\text{3000 }\Omega }{\text{300 }\Omega +2\text{00 }\Omega +150\Omega }\\ =\frac{3000\Omega }{\text{650 }\Omega }\\ =5\Omega \end{array}$

The current passing through a parallel circuit is divided among each resistor.

The expression for current passing through each resistor is,

$I=I_1+I_2+I_3$ (1)

Here,

$I_1,I_2,I_3$ are the current passing through each resistor

Write the expression for current in terms of voltage and resistance in a parallel circuit.

$I=\frac{V}{R}$

Here,

$V$ is the dc voltage source across the parallel circuit

Substitute $\frac{V}{R_1}$ for $I_1$ , $\frac{V}{R_2}$ for $I_2$ and $\frac{V}{R_3}$ for $I_3$ in equation (1).

$I=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}$

Substitute $50\text{ V}$ for $V$, $10\Omega$ for $R_1$, $15\Omega$ for $R_2$ and $\text{20 }\Omega$ for $R_3$ to get $I$.

$\begin{array}{c}I=\frac{50\text{ V}}{10\Omega }+\frac{50\text{ V}}{15\Omega }+\frac{50\text{ V}}{\text{20 }\Omega }\\ =\left(50\text{ V}\right)\left(\frac{1}{10\Omega }+\frac{1}{15\Omega }+\frac{1}{20\Omega }\right)\\ =\left(50\text{ V}\right)\left(\frac{\left(\text{15 }\Omega \right)\left(\text{20 }\Omega \right)+\left(\text{10 }\Omega \right)\left(\text{20 }\Omega \right)+\left(\text{10 }\Omega \right)\left(\text{15 }\Omega \right)}{\left(\text{10 }\Omega \right)\left(\text{15 }\Omega \right)\left(\text{20 }\Omega \right)}\right)\\ =\left(50\text{ V}\right)\left(\frac{650\Omega }{\text{3000 }\Omega }\right)\end{array}$

Further, solve the equation.

$\begin{array}{c}I=\frac{50\text{ V}}{5\Omega }\\ =10\text{ A}\end{array}$

Conclusion:

Therefore, the current passing through the resistors in a parallel circuit is $10\text{ A}$.

(b)

To determine

The power expended in the parallel circuit.

Answer to Problem 22E

The power expended in the parallel circuit is $500\text{ W}$.

Explanation of Solution

Given info: First resistor having resistance $10\Omega$, second resistor having resistance $\text{15 }\Omega$ , third resistor having resistance $\text{20 }\Omega$ and dc voltage source is $\text{50 V}$.

Explanation:

The expression for power in terms of voltage and resistance is,

$P=VI$

Substitute $12\text{ V}$ for $V$ and $0.8\text{ A}$ for $I$ in the above expression.

$\begin{array}{c}P=\left(50\text{ V}\right)\left(10\text{ A}\right)\\ =500\text{ W}\end{array}$

Conclusion:

Therefore, the power expended in the parallel circuit is $500\text{ W}$.