Chapter 23.1, Problem 15PP

a.

To determine

The equivalent resistance of the parallel circuit.

Answer to Problem 15PP

The equivalent resistance of the network is $\text{4}\text{.28 }\Omega$.

Explanation of Solution

Given:

A parallel circuit consisting of two resistors of value 15 Ω each and third resistor being 10 Ω are connected 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,

  $\begin{array}{c}\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{15}+\frac{1}{15}\\ R_{eq}=4.28\Omega \end{array}$

With reference to the case when all the resistance values were 15 Ω, the equivalent resistance decreases slightly when one of the resistances is decreased to 10 Ω.

Conclusion:

The equivalent resistance of the circuit is $R_{eq}=4.28\Omega$ .

b.

To determine

The current in the circuit

Answer to Problem 15PP

The current through the circuit is $\text{7 A}$

Explanation of Solution

Given:

A parallel circuit consisting of two resistors of value 15 Ω each and third resistor being 10 Ω are connected 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

  $\begin{array}{c}I=\frac{V}{R_{eq}}\\ =\frac{30}{4.28}\\ =7\text{ A}\end{array}$

With reference to the case when all the resistance values were 15 Ω, the total current increases by 1 A when one of the resistances is decreased to 10 Ω.

Conclusion:

The current through the circuit is 7 A.

c.

To determine

To explain: The effect of change in current through 15 Ω due to a given change in resistance of the circuit.

Answer to Problem 15PP

The current flowing through 15 Ω branch does not change and it remains as 2 A.

Explanation of Solution

Given:

A parallel circuit consisting of two resistors of value 15 Ω each and third resistor being 10 Ω are connected 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, consider the first branch consisting of 10 Ω, the current flowing through it is,

  $\begin{array}{c}{I}_{10\Omega }=\frac{30}{10}\\ =3\text{ A}\end{array}$

Now, the current through the two branches with 15 Ω are equal and can be calculated as,

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

It can be observed that, the current through the 15 Ω remains the same though the total current shows an increase in value.

Conclusion:

The current in 15Ω remains the same as when the circuit has all resistance values as 15 Ω.