Chapter 29, Problem 72A

To determine

To find the resistors to be connected to the two LEDs that glow with different colors.

Answer to Problem 72A

Resistors of approximately $180\Omega \text{ and 160 }\Omega$ are to be used for the two given LEDs.

Explanation of Solution

Given data:

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

Current through each LED, $I=0.040\text{ A}$

The current versus potential difference characteristic of the two LEDs

  

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, $V_b=9.0\text{ V}$

Current through each LED, $I=0.040\text{ A}$

Consider two resistors $R_1$ for first LED and $R_2$ for second LED.

For the first LED,

Potential difference from the given graph is $1.75\text{ V}$ at $I=0.040\text{ A}$

Calculate resistance using the formula

  $\begin{array}{l}V=IR\\ R=\frac{V}{I}\\ R=\frac{V_b-V_D}{I}\end{array}$

Substitute the values $V_b=9.0\text{ V}$ , $V_D=1.75\text{ V}$ and $I=0.040\text{ A}$

  $\begin{array}{l}{R}_1=\frac{9.0\text{ eV-1}\text{.75 eV}}{0.040\text{ A}}\\ R_1=181.25\Omega \end{array}$

For the second LED,

Potential difference from the given graph is $2.5\text{ V}$ at $I=0.040\text{ A}$

Calculate resistance using the formula

  $\begin{array}{l}V=IR\\ R=\frac{V}{I}\\ R=\frac{V_b-V_D}{I}\end{array}$

Substitute the values $V_b=9.0\text{ V}$ , $V_D=2.5\text{ V}$ and $I=0.040\text{ A}$

  $\begin{array}{l}{R}_2=\frac{9.0\text{ eV-2}\text{.5 eV}}{0.040\text{ A}}\\ R_2=162.5\Omega \end{array}$

Resistors of approximately $180\Omega \text{ and 160 }\Omega$ are to be used for the two given LEDs.

Conclusion:

Thus,Resistors of approximately $180\Omega \text{ and 160 }\Omega$ are to be used for the two given LEDs.