Chapter 29, Problem 50A

a.

To determine

To calculate: Current through the LED.

Answer to Problem 50A

Current through the LED = $\underset{\_}{\text{0}\text{.02 A}}$

Explanation of Solution

Given:

The following figure:

  

Battery’s voltage = $\text{6}\text{.0 V}$

Resistance = $\text{240 }\Omega$

Potential difference across LED = $1.2\text{ V}$

Formula used:

The power supply voltage is given by,

  $V_{\text{b}}=IR+V_{\text{d}}$

Where $I$ is the current, $R$ is the resistance and $V_{\text{d}}$ is the potential difference.

Calculation:

Current through the LED is calculated as:

  $\begin{array}{l}I=\frac{V_{\text{b}}-V_{\text{d}}}{R}\\ =\frac{6.0\text{ V}-1.2\text{ V}}{240\Omega }\\ =0.02\text{ A}\end{array}$

Conclusion:

Current through the LED = $\underset{\_}{\text{0}\text{.02 A}}$

b.

To determine

To calculate: Current through the resistor.

Answer to Problem 50A

Current through the resistor = $\underset{\_}{\text{0}\text{.02 A}}$

Explanation of Solution

Given:

The following figure:

  

Battery’s voltage = $\text{6}\text{.0 V}$

Resistance = $\text{240 }\Omega$

Potential difference across LED = $1.2\text{ V}$

Potential difference across the resistor is the difference between battery’s potential difference and the LED’s potential difference.

Formula used:

The power supply voltage is given by,

  $V_{\text{b}}=IR+V_{\text{d}}$

Where $I$ is the current, $R$ is the resistance and $V_{\text{d}}$ is the potential difference.

Calculation:

Current through the resistor is calculated as:

  $\begin{array}{l}I=\frac{V_{\text{b}}-V_{\text{d}}}{R}\\ =\frac{6.0\text{ V}-1.2\text{ V}}{240\Omega }\\ =0.02\text{ A}\end{array}$

Conclusion:

Current through the resistor = $\underset{\_}{\text{0}\text{.02 A}}$