Chapter 8, Problem 1VC

To determine

The current passing through each resistor and a parallel circuit combination.

Answer to Problem 1VC

The current passing through each resistors are $1\text{ A}$ for $I_a$, $2\text{ A}$ for $I_b$ and $2\text{ A}$ for $I_c$ and the total current passing through a parallel circuit is $5\text{ A}$ for $I_d$.

Explanation of Solution

Given info: First resistor having resistance $\text{12 }\Omega$, second resistor having resistance $\text{6 }\Omega$ , third resistor having resistance $\text{6 }\Omega$ and dc voltage source is $\text{12 V}$.

Explanation:

The figure represents a parallel circuit arrangement. In a parallel circuit three 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 $\text{12 }\Omega$ for $R_1$ and $\text{6 }\Omega$ for $R_2$ and $\text{6 }\Omega$ for $R_3$.

$\begin{array}{c}{R}_P=\frac{\left(\text{12 }\Omega \right)\left(\text{6 }\Omega \right)\left(\text{6 }\Omega \right)}{\left(\text{6 }\Omega \right)\left(\text{6 }\Omega \right)+\left(\text{12 }\Omega \right)\left(\text{6 }\Omega \right)+\left(\text{6 }\Omega \right)\left(\text{12 }\Omega \right)}\\ =\frac{\text{432 }\Omega }{36\Omega +\text{72 }\Omega +72\Omega }\\ =\frac{432\Omega }{\text{180 }\Omega }\\ =2.4\Omega \end{array}$

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

The expression for the total current passing through each resistor is,

$I_d=I_a+I_b+I_c$ (1)

Here,

$I_a,I_b,I_c$ are the current passings through each resistor

$I_d$ is the total current flowing through a parallel circuit

Write the expression for current in terms of voltage and resistance in a resistor having resistance $R_1$.

$I_a=\frac{V}{R_1}$

Here,

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

Substitute $12\text{ V}$ for $V$ and $12\Omega$ for $R_1$ to get $I_a$.

$\begin{array}{c}{I}_a=\frac{12\text{ V}}{12\Omega }\\ =1\text{ A}\end{array}$

Write the expression for current in terms of voltage and resistance in a resistor having resistance $R_2$.

$I_b=\frac{V}{R_2}$

Substitute $12\text{ V}$ for $V$ and $6\Omega$ for $R_2$ to get $I_b$.

$\begin{array}{c}{I}_b=\frac{12\text{ V}}{6\Omega }\\ =2\text{ A}\end{array}$

Write the expression for current in terms of voltage and resistance in a resistor having resistance $R_3$.

$I_c=\frac{V}{R_3}$

Substitute $12\text{ V}$ for $V$ and $6\Omega$ for $R_3$ to get $I_c$.

$\begin{array}{c}{I}_c=\frac{12\text{ V}}{6\Omega }\\ =2\text{ A}\end{array}$

The expression for the total current passing through each resistor is,

$I_d=I_a+I_b+I_c$ (2)

Substitute $\frac{V}{R_1}$ for $I_a$ , $\frac{V}{R_2}$ for $I_b$ and $\frac{V}{R_3}$ for $I_c$ in equation (2).

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

Substitute $12\text{ V}$ for $V$, $12\Omega$ for $R_1$, $6\Omega$ for $R_2$ and $\text{6 }\Omega$ for $R_3$ to get $I_d$.

$\begin{array}{c}{I}_d=\frac{12\text{ V}}{12\Omega }+\frac{12\text{ V}}{\text{6 }\Omega }+\frac{12\text{ V}}{\text{6 }\Omega }\\ =\left(12\text{ V}\right)\left(\frac{1}{12\Omega }+\frac{1}{6\Omega }+\frac{1}{6\Omega }\right)\\ =\left(12\text{ V}\right)\left(\frac{\left(\text{6 }\Omega \right)\left(\text{6 }\Omega \right)+\left(\text{12 }\Omega \right)\left(\text{6 }\Omega \right)+\left(\text{12 }\Omega \right)\left(\text{6 }\Omega \right)}{\left(\text{12 }\Omega \right)\left(\text{6 }\Omega \right)\left(\text{6 }\Omega \right)}\right)\end{array}$

Further solve the equation.

$\begin{array}{c}{I}_d=\left(12\text{ V}\right)\left(\frac{180\Omega }{\text{432 }\Omega }\right)\\ =\frac{2160\text{ V}}{432\Omega }\\ =5\text{ A}\end{array}$

Conclusion:

Therefore, the current passing through each resistor is $1\text{ A}$ for $I_a$, $2\text{ A}$ for $I_b$ and $2\text{ A}$ for $I_c$ and the total current passing through a parallel circuit is $5\text{ A}$ for $I_d$.