Chapter 22, Problem 70A

a.

To determine

The resistance of the diode when the potential difference across it is $0.70\text{V}$ .

Answer to Problem 70A

  $35\Omega$

Explanation of Solution

Given:

The current versus voltage graph.

The voltage across the diode, at which resistance is to be calculated, $V=+0.7\text{V}$

Formula used:

The electric resistance is given by,

  $R=\frac{V}{I}$

Where, I is the current and V is the potential difference.

Calculation:

From the given graph, when the voltage is $+0.7\text{V}$ , the current through the diode is, $I=0.02\text{A}$ .

Then, using the above formula and the given values, the resistance of the diode will be,

  $\begin{array}{c}R=\frac{V}{I}\\ =\frac{0.7}{0.02}\\ =35\Omega \end{array}$

Conclusion:

Hence, theresistance of the diode when the potential difference across it is $0.70\text{V}$ is $35\Omega$ .

b.

To determine

The resistance of the diode when the potential difference across it is $0.60\text{V}$ .

Answer to Problem 70A

  $100\Omega$

Explanation of Solution

Given:

The current versus voltage graph.

The voltage across the diode, at which resistance is to be calculated, $V=+0.6\text{V}$

Formula used:

The electric resistance is given by,

  $R=\frac{V}{I}$

Where, I is the current and V is the potential difference.

Calculation:

From the given graph, when the voltage is $+0.6\text{V}$ , the current through the diode is, $I\approx 0.006\text{A}$

Then, using the above formula and the given values, the resistance of the diode will be,

  $\begin{array}{c}R=\frac{V}{I}\\ =\frac{0.6}{0.006}\\ =100\Omega \end{array}$

Conclusion:

Hence, theresistance of the diode when the potential difference across it is $0.60\text{V}$ , is $100\Omega$ .