Chapter 2, Problem 2.1QAP

Interpretation Introduction

Interpretation:

A suitable combination of the resistors with the indicated voltage needs to be determined.

Concept Introduction :

The resistors in series from a voltage divider in that fraction of the total voltage appear across each resistor.

Answer to Problem 2.1QAP

A suitable combination of the resistors that would give the indicated voltages are $R_1=500\Omega ,R_2=2k\Omega ,R_3=2.5k\Omega .$ .

Explanation of Solution

Consider the diagram, all the resistors ( $R_1,R_2,\text{and }{R}_3$ ) are in series.

The resistors in series from a voltage divider in that fraction of the total voltage appear across each resistor.

Applying Ohm’s law,from A to B, the voltage across the resistor $R_1$ is:

  $V_1=I_1{R}_1$

The fraction of the total voltage appears across resistor $R_1$ is,

  $\begin{array}{l}\frac{V_1}{V}=\frac{I{R}_1}{I\left(R_1+R_2+R_3\right)}\\ \\ =\frac{R_1}{\left(R_1+R_2+R_3\right)}\\ \\ =\frac{R_1}{R_s}\end{array}$

Here

  $R_s=R_1+R_2+R_3$

Given

  $V=10.0V,V_1=1.0V,V_2=4.0V$

Therefore,

  $\begin{array}{l}\frac{V_1}{V}=\frac{1.0V}{10.0V}=\frac{1}{10}=\frac{R_1}{R_s}\\ \\ R_s=10R_1\end{array}$

Similarly,

  $\begin{array}{l}\frac{V_2}{V}=\frac{R_2}{R_s}\\ \\ \frac{4.0V}{10.0V}=\frac{R_2}{R_s}\\ \\ R_s=\frac{10}{4}{R}_2\end{array}$

Substitute the value $R_s=10R_1$

Therefore,

  $\begin{array}{l}10R_1=\frac{10}{4}{R}_2\\ \\ \frac{R_1}{R_2}=\frac{1}{4}\end{array}$

Here, $R_1=500\Omega ,R_2=2k\Omega =2000\Omega$ .

Hence, it is shown that

:

  $\frac{R_1}{R_2}=\frac{500}{2000}=\frac{1}{4}$

In the case of series resistors, the value of $R_s$ is:

  $R_s=R_1+R_2+R_3$

  $\begin{array}{l}{R}_s=500+2000+R_3\\ \\ \frac{V_1}{V_s}=\frac{R_1}{R_s}\end{array}$

Or,

  $\begin{array}{l}\frac{1}{10}=\frac{500}{500+2000+R_3}\\ \\ R_3=2500\Omega =2.5k\Omega \end{array}$

Interpretation Introduction

Interpretation:

The IR drop across $R_3$ needs to be determined.

Concept Introduction :

According to the Kirchhoff’s law, the algebraic sum of the voltage around the closed path is zero.

Answer to Problem 2.1QAP

IR drop across $R_3$ is 5.0V.

Explanation of Solution

According to the Kirchhoff’s law, the algebraic sum of the voltage around the closed path is zero.

It is represented as

  $V=V_1+V_2+V_3$

Given,

  $V=10.0V,V_1=1.0V,V_2=4.0V$

  $\begin{array}{l}10.0V=1.0V+4.0V+V_3\\ \\ V_3=5.0\end{array}$

IR drop across $R_3$ is 5.0V.

Interpretation Introduction

Interpretation:

The current drawn from the source needs to be determined.

Concept Introduction : According to the Ohms law $V=IR$ .

Here, V is voltage, I is current, and R is resistance.

Answer to Problem 2.1QAP

Current drawn from the source is 0.002A.

Explanation of Solution

According to Ohm’s law,

  $V=IR$

Consider the series resistors,

  $\begin{array}{l}R=R_1+R_2+R_3\\ \\ =500\Omega +2000\Omega +2500\Omega \\ \\ =5000\Omega \\ \\ I=\frac{V}{R}\\ \\ =\frac{10V}{5000\Omega }=0.002A\end{array}$

Thus, the current drawn from the source is 0.002A.

Interpretation Introduction

Interpretation:

The dissipated power needs to be calculated.

Concept Introduction:

The power dissipated in the circuit is the product of current and potential difference across the element.

Answer to Problem 2.1QAP

The power dissipated by the circuit is 0.02W.

Explanation of Solution

Consider the power is:

  $\begin{array}{l}P=VI\\ \\ P=0.00\text{2 A×10}\text{.0 V}\\ \\ =0.\text{02 W}\end{array}$

Hence, the dissipated power is 0.02W.