Chapter 23.1, Problem 16PP

a.

To determine

The equivalent resistance of the parallel circuit.

Answer to Problem 16PP

  $20\Omega$

Explanation of Solution

Given:

The potential applied, $V=12.0\text{V}$

The first resistor, $R_1=120.0\Omega$

The second resistor, $R_2=60\Omega$

The third resistor, $R_3=40.0\Omega$

Formula used:

In parallel combination of resistors, the net resistance is,

  $\frac{1}{R_n}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Calculation:

Using above formula, the net resistance will be,

  $\begin{array}{l}\frac{1}{R_n}=\frac{1}{120}+\frac{1}{60}+\frac{1}{40}\\ R_n=20\Omega \end{array}$

Conclusion:

Hence, the net resistance of the parallel combination is $20\Omega$ .

b.

To determine

The net current through the parallel circuit.

Answer to Problem 16PP

  $20\Omega$

Explanation of Solution

Given:

The potential applied, $V=12.0\text{V}$

The first resistor, $R_1=120.0\Omega$

The second resistor, $R_2=60\Omega$

The third resistor, $R_3=40.0\Omega$

From part (a), the net resistance, $R_n=20.0\Omega$

Formula used:

Ohm’s law,

  $V=IR$

Where,V is the voltage, I is the current and R is the resistance.

Calculation:

Using above formula, the net current through the circuit will be,

  $\begin{array}{l}V=I_n{R}_n\\ I_n=0.6\text{A}\end{array}$

Conclusion:

Hence, the net current through the parallel combination is $0.6\text{A}$ .

c.

To determine

The current through each resistor.

Answer to Problem 16PP

  $I_1=0.1\text{A}$ , $I_2=0.2\text{A}$ , and $I_3=0.3\text{A}$ .

Explanation of Solution

Given:

The potential applied, $V=12.0\text{V}$

The first resistor, $R_1=120.0\Omega$

The second resistor, $R_2=60\Omega$

The third resistor, $R_3=40.0\Omega$

From part (a), the net resistance, $R_n=20.0\Omega$

Formula used:

Ohm’s law,

  $V=IR$

Where,V is the voltage, I is the current and R is the resistance.

Calculation:

It is known that in parallel, potential remains same. Thus, the current through the first resistor is,

  $\begin{array}{c}V=I_1{R}_1\\ I_1=\frac{12}{120}\\ =0.1\text{A}\end{array}$

Similarly, the current through the second resistor is,

  $\begin{array}{c}V=I_2{R}_2\\ I_2=\frac{12}{60}\\ =0.2\text{A}\end{array}$

And, the current through the third resistor is,

  $\begin{array}{c}V=I_3{R}_3\\ I_3=\frac{12}{40}\\ =0.3\text{A}\end{array}$

Conclusion:

Hence, it can be observed that the net current through the circuit is equal to the sum of the current through each resistor.