Chapter 23.1, Problem 5PP

To determine

The resistance across the given each resistors.

Answer to Problem 5PP

  $V_1=28.08\text{V}$ , $V_2=35.1\text{V}$ ,and $V_3=11.7\text{V}$ .

Explanation of Solution

Given:

The potential applied, $V=75\text{V}$

The first resistor, $R_1=12\Omega$

The second resistor, $R_2=15\Omega$

The third resistor, $R_3=5\Omega$

Formula used:

In series combination of resistors, the net resistance is,

  $R_n=R_1+R_2+R_3$

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 resistance will be,

  $\begin{array}{c}{R}_n=R_1+R_2+R_3\\ =32\Omega \end{array}$

Then, the net current through the combination,

  $\begin{array}{c}I=\frac{V}{R_n}\\ =\frac{75}{32}\\ =2.34\text{A}\end{array}$

Same current flows in series combination, thus potential across first resistor is,

  $\begin{array}{c}{V}_1=I{R}_1\\ =28.08\text{V}\end{array}$

Similarly, the potential across second resistor is,

  $\begin{array}{c}{V}_2=I{R}_2\\ =35.1\text{V}\end{array}$

And, the potential across third resistor is,

  $\begin{array}{c}{V}_3=I{R}_3\\ =11.7\text{V}\end{array}$

It can be seen that,

  $\begin{array}{c}{V}_1+V_2+V_3=\left(28.08+35.1+11.7\right)\text{V}\\ \approx 75\text{V}\end{array}$

Which is equal to the applied voltage.

Conclusion:

Hence, the voltage applied is equal to the sum of the voltage across each resistor.