Chapter 1, Problem 1.1P

(a)

To determine

The value of current I by using Ohm’s Law.

Answer to Problem 1.1P

  $I=5\text{mA}$

Explanation of Solution

Given information:

  $\begin{array}{l}V=5\text{V}\\ R=1\text{kΩ}\end{array}$

Calculation:

The expression of Ohm’s law can be written as

  $V=IR$

Here, V = voltage

I = current

R = resistance.

Putting the given values in the equation of Ohm’s Law,

  $\begin{array}{l}V=IR\\ 5=I\left(1\times {10}^3\right)\\ I=\frac{5}{1\times {10}^3}\\ I=5\times {\text{10}}^{-3}\text{A}\\ I=5\text{mA}\end{array}$

(b)

To determine

The value of resistanceR by using Ohm’s Law.

Answer to Problem 1.1P

  $R=5\text{kA}$

Explanation of Solution

Given information:

  $\begin{array}{l}I=1\text{mA}\\ V=5\text{V}\end{array}$

Calculation:

The expression of Ohm’s law can be written as

  $V=IR$

Here, V = voltage

I = current

R = resistance.

Putting the given values in the equation of Ohm’s Law,

  $\begin{array}{l}V=IR\\ 5=R\left(1\times {10}^{-3}\right)\\ R=\frac{5}{1\times {10}^{-3}}\\ R=5\times {\text{10}}^3\text{A}\\ R=5\text{kA}\end{array}$

(c)

To determine

The value of voltage V by using Ohm’s Law.

Answer to Problem 1.1P

  $V=1\text{V}$

Explanation of Solution

Given information:

  $\begin{array}{l}I=0.1\text{mA}\\ R=10\text{k}\Omega \end{array}$

Calculation:

The expression of Ohm’s law can be written as

  $V=IR$

Here, V = voltage

I = current

R = resistance.

Putting the given values in the equation of Ohm’s Law,

  $\begin{array}{l}V=IR\\ V=\left(10\times {10}^3\right)\left(0.1\times {10}^{-3}\right)\\ V=1\text{V}\end{array}$

(d)

To determine

The value of current I by using Ohm’s Law.

Answer to Problem 1.1P

  $I=10\text{mA}$

Explanation of Solution

Given information:

  $\begin{array}{l}V=1\text{V}\\ R=10\text{0}\Omega \end{array}$

Calculation:

The expression of Ohm’s law can be written as

  $V=IR$

Here, V = voltage

I = current

R = resistance.

Putting the given values in the equation of Ohm’s Law,

  $\begin{array}{l}V=IR\\ 1=I\left(100\right)\\ I=\frac{1}{100}\text{A}\\ I=10\text{mA}\end{array}$