Chapter 23, Problem 78A

a.

To determine

To rank: the current through each resistor from least to greatest.

Answer to Problem 78A

  $I_{30\Omega }=I_{20\Omega }=I_{10\Omega }=I_{40\Omega }<I_{30\Omega }=I_{20\Omega }$

Explanation of Solution

Given:

The given figure is,

  

Formula used:

Ohm’s law is given as,

  $V=IR$

Where, $V$ is voltage across load, $I$ is current through load and $R$ is load resistance.

Calculation:

For the given circuit,

  

Here, equivalent resistance is

  $\begin{array}{l}{R}_{\text{eq}}=25+\left(30+20\right)\left|\right|\left(10+40\right)\\ =25+\left(50\left|\right|50\right)\\ =25+25\\ =50\Omega \end{array}$

So, the total current

  $I=\frac{V}{R}=\frac{25}{50}=0.5\text{A}$

Now,

  $\begin{array}{l}{I}_1=0.5\times \frac{50}{100}=0.25\text{A}\text{and}\\ I_2=0.5\times \frac{50}{100}=0.25\text{A}\end{array}$

So, current in resistors are

  $\begin{array}{l}{I}_{25\Omega }=I=0.5\text{A}\\ I_{30\Omega }=I_{20\Omega }=I_1=0.25\text{A}\\ I_{10\Omega }=I_{40\Omega }=I_2=0.25\text{A}\end{array}$

Conclusion:

Therefore, current ranking in resistors is $I_{30\Omega }=I_{20\Omega }=I_{10\Omega }=I_{40\Omega }<I_{30\Omega }=I_{20\Omega }$ .

b.

To determine

To rank: the voltage across each resistor from least to greatest.

Answer to Problem 78A

  $V_{10\Omega }<V_{\text{20}\Omega }<V_{30\Omega }<V_{40\Omega }<V_{\text{25}\Omega }$

Explanation of Solution

The given figure is,

  

Formula used:

Ohm’s law is given as,

  $V=IR$

Where, $V$ is voltage across load, $I$ is current through load and $R$ is load resistance.

Calculation:

For the given circuit,

  

Here, from the previous subpart the current in resistors are,

  $\begin{array}{l}{I}_{25\Omega }=I=0.5\text{A}\\ I_{30\Omega }=I_{20\Omega }=I_1=0.25\text{A}\\ I_{10\Omega }=I_{40\Omega }=I_2=0.25\text{A}\end{array}$

So, voltage across each resistor is

  $\begin{array}{l}{V}_{\text{25}\Omega }=25\times I_{25\Omega }=25\times 0.5=12.5\text{V}\\ V_{30\Omega }=30\times I_{30\Omega }=30\times 0.25=7.5\text{V}\\ V_{\text{20}\Omega }=20\times I_{20\Omega }=20\times 0.25=5\text{V}\\ V_{10\Omega }=10\times I_{10\Omega }=10\times 0.25=2.5\text{V}\\ V_{40\Omega }=40\times I_{40\Omega }=40\times 0.25=10\text{V}\end{array}$

Conclusion:

Therefore, voltage ranking of resistor is $V_{10\Omega }<V_{\text{20}\Omega }<V_{30\Omega }<V_{40\Omega }<V_{\text{25}\Omega }$