Chapter 23.1, Problem 14PP

a.

To determine

The equivalent resistance of the parallel circuit.

Answer to Problem 14PP

The equivalent resistance of the network is $\text{5 }\Omega$.

Explanation of Solution

Given:

Three resistors of value 15 Ω each are connected in parallel across a 30-V battery.

Formula used:

The equivalent resistance of a parallel branch of two resistances is given by,

  $\frac{1}{R_{eq}}=\frac{1}{R_A}+\frac{1}{R_B}$ …… (1)

Calculation:

Consider the circuit shown in Figure 1.

  

Figure 1

Here, the equivalent resistance of the parallel branches can be calculated using (1) as,

  $\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{15}+\frac{1}{15}+\frac{1}{15}\\ R_{eq}=5\Omega \end{array}$

Conclusion:

The equivalent resistance of the circuit is $R_{eq}=5\Omega$ .

b.

To determine

The current in the circuit.

Answer to Problem 14PP

The current through the circuit is $\text{6 A}$.

Explanation of Solution

Given:

Three resistors of value 15 Ω each are connected in parallel across a 30-V battery.

Formula used:

For the resistance of the circuit and supply voltage, the current can be calculated using the Ohm’s law as,

  $\begin{array}{l}V=IR\\ I=\frac{V}{R}\end{array}$ …… (2)

Calculation:

The current flowing through the circuit can be calculated by using the relation in (2), by substituting the total resistance obtained in the part a and the supply voltage V=30 V, as

  $\begin{array}{c}I=\frac{V}{R_{eq}}\\ =\frac{30}{5}\\ =6\text{ A}\end{array}$

Conclusion:

The current through the circuit is 6 A.

c.

To determine

The current flowing in each branch of the circuit.

Answer to Problem 14PP

The current flowing through each of the branch is $I_1=I_2=I_3=2\text{A}$.

Explanation of Solution

Given:

Three resistors of value 15 Ω each are connected in parallel across a 30-V battery.

Formula used:

The voltage in the each branch in a parallel circuit is equal. The current flowing through each of the branch is expressed in terms of the supply voltage and their individual resistance values as,

  $I=\frac{V}{R}$

Calculation:

Here, since all the three branches have the same resistance values, the branch currents are the same. It can be calculated as,

  $\begin{array}{c}I=\frac{V}{R}\\ =\frac{30}{15}\\ =2\text{ A}\end{array}$

Thus, the three branches has current of 2 A respectively.

Conclusion:

The current in each of the parallel branches is 2 A.