Chapter 29, Problem 73A

To determine

To find the resistor to be connected to the two LEDs,if connected in series.

Answer to Problem 73A

A resistor of approximately $140\Omega$ is to be used for the two LEDs in series.

Explanation of Solution

Given data:

Supply voltage, $V_b=9.0\text{ V}$

Desired current through, $I=0.035\text{ A}$

Potential drop across first LED, $V_{D1}=1.75\text{ V}$

Potential drop across second LED, $V_{D2}=2.5\text{ V}$

Formula used:

From ohm’s law:

  $V=IR$

  $V$ represents the voltage across the resistor $R$ and $I$ through the resistor.

Calculation:

As per problem

Supply voltage, $V_b=9.0\text{ V}$

Potential drop across first LED, $V_{D1}=1.75\text{ V}$

Potential drop across second LED, $V_{D2}=2.5\text{ V}$

The voltage drop across the two LEDs combines is $V_{D1}+V_{D2}=1.75\text{ V+2}\text{.5 V=4}\text{.25 V}$

Rearrange the stated formula for $R$

  $R=\frac{V_B\text{-(}{V}_{D1}+V_{D2})}{I}$

Substitute the values $V_b=9.0\text{ V}$ , $V_{D1}+V_{D2}=\text{4}\text{.25 V}$ and $I=0.035\text{ A}$

  $\begin{array}{l}R=\frac{9.0\text{ V-4}\text{.25 V}}{0.035\text{ A}}\\ R=135.7\Omega \end{array}$

A resistor of approximately $140\Omega$ is to be used for the two LEDs in series.

Conclusion:

Thus, a resistor of approximately $140\Omega$ is to be used for the two LEDs in series.