Chapter 29, Problem 53A

a.

To determine

To calculate: Base current.

Answer to Problem 53A

Base current = $\underset{\_}{2.33\times {10}^{-5}\text{ A}}$

Explanation of Solution

Given:

The following figure:

  

The switch shown in the figure is on.

Voltage drop across base-emitter junction = $\text{0}\text{.70 V}$

Current gain from base-collector junction = $\text{220 A}$

Formula used:

Current is given by,

  $I=\frac{V_{\text{b}}-V_{\text{d}}}{R}$

Where y $V_{\text{b}}$ is the power supply voltage, 1 $V_{\text{d}}$ is voltage drop and 3

  $R$ is resistance.

Calculation:

Base current is calculated as:

  $\begin{array}{l}I=\frac{V_{\text{b}}-V_{\text{d}}}{R}\\ =\frac{(3.5-0.70)\text{ V}}{120000}\\ =2.33\times {10}^{-5}\text{ A}\end{array}$

Conclusion:

Base current = $\underset{\_}{2.33\times {10}^{-5}\text{ A}}$

b.

To determine

To calculate: Collector current.

Answer to Problem 53A

Collector current = $\underset{\_}{\text{5}\text{.126}\times {10}^{-3}\text{ A}}$

Explanation of Solution

Given:

The following figure:

  

The switch shown in the figure is on.

Voltage drop across base-emitter junction = $\text{0}\text{.70 V}$

Current gain from base-collector junction = $\text{220 A}$

Formula used:

Current gain is given by,

  $\beta =\frac{I_{\text{C}}}{I_{\text{B}}}$

  $$

Where $I_{\text{C}}$ is collector current and $I_{\text{B}}$ is base current.

Calculation:

Collector current is calculated as:

  $\begin{array}{l}{I}_{\text{C}}=\beta \times I_{\text{B}}\\ =220\text{ A}\times (2.33\times {10}^{-5})\text{ A}\\ \text{ =5}\text{.126}\times {10}^{-3}\text{ A}\end{array}$

Conclusion:

Collector current = $\underset{\_}{\text{5}\text{.126}\times {10}^{-3}\text{ A}}$

c.

To determine

To calculate: Voltmeter reading.

Answer to Problem 53A

Voltmeter reading = $\underset{\_}{\text{7}\text{.3 V}}$

Explanation of Solution

Given:

The following figure:

  

The switch shown in the figure is on.

Voltage drop across base-emitter junction = $\text{0}\text{.70 V}$

Current gain from base-collector junction = $\text{220 A}$

Formula used:

Voltage is given by,

  $V=IR$

Where $I$ is current and $R$ is resistance.

The voltmeter reading is given by,

  $V_{\text{transistor}}=V_{\text{battery}}-V_{\text{resistor}}$

Calculation:

Voltage across $\text{1500}\Omega$ resistor is calculated as:

  $\begin{array}{l}V=IR\\ =(\text{5}\text{.126}\times {\text{10}}^{{}^{-3}}\text{A})(\text{1500}\Omega )\\ =7.689\text{ V}\end{array}$

Voltmeter reading is calculated as,

  $\begin{array}{l}{V}_{\text{transistor}}=V_{\text{battery}}-V_{\text{resistor}}\\ =15\text{ V}-7.689\text{ V}\\ \text{ =7}\text{.3 V}\end{array}$

Conclusion:

Voltmeter reading = $\underset{\_}{\text{7}\text{.3 V}}$