Chapter 29, Problem 71A

(a)

To determine

To find the voltmeter reading.

Answer to Problem 71A

The voltmeter reading is $0.7\text{ V}$ .

Explanation of Solution

Given:

The figure:

  

Calculation:

As per problem,

Consider the given figure

  

The diodes are forward-biased (positive terminal of the battery is connected to the p-side and d the negative terminal is connected to the n-side)

The voltage drop across a silicon diode is $0.7\text{ V}$ when it is forward biased.

Hence, the voltmeter connected across the parallel diode circuit reads $0.7\text{ V}$ .

Conclusion:

The voltmeter reading is $0.7\text{ V}$ .

(b)

To determine

To find the reading of ammeter ${\text{A}}_1$ .

Answer to Problem 71A

The reading of the ammeter ${\text{A}}_1$ is 1 $0\text{ A}$ .

Explanation of Solution

Given:

The figure:

  

Calculation:

The voltage drop across the two diodes in series is $0.7\text{ V}$ .

Such small voltage is not enough to turn on the two diodes.

Therefore, current through the diodes are $0\text{ A}$ and so the ammeter ${\text{A}}_1$ reads $0\text{ A}$ .

Conclusion:

The reading of the ammeter ${\text{A}}_1$ is $0\text{ A}$ .

(c)

To determine

To find the reading of ammeter ${\text{A}}_2$ .

Answer to Problem 71A

The reading of the ammeter ${\text{A}}_2$ is $\text{42 mA}$ .

Explanation of Solution

Given:

The figure:

  

Formula used:

From ohm’s law:

  $V=IR$

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

Calculation:

As per the circuit given,

The voltage drop across the diode to which ${\text{A}}_2$ is connected is $V_d=0.7\text{ V}$ .

Supply voltage, $V_b=10.0\text{ V}$

The resistance of the resistor in the circuit, $R=220\Omega$

Use the above stated formula to get current

  $\begin{array}{l}I=\frac{V}{R}\\ I=\frac{V_b-V_D}{R}\\ I=\frac{10.0\text{ V-0}\text{.7 V}}{220\Omega }\\ I=0.042\text{ A}\\ \text{I=42 mA}\end{array}$

Conclusion:

Thus, the reading of the ammeter ${\text{A}}_2$ is $\text{42 mA}$ .