Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Table of contents
Textbook solutions
All Solutions
Page 786: Practice Problems
Exercise 22
Step 1
1 of 4
In this problem, we are given a battery that powers a diode with a given voltage in a circuit of known current, and resistance. We should find the battery voltage.
Step 2
2 of 4
The equation that describes this process is Ohm’s law applied on the diode circuit
$$
V_b=IR+V_d
$$
Step 3
3 of 4
center{Now, we can insert the given values to have that}
[V_b=2.5times 10^{-3}times 470 +0.7=boxed{1.9 textrm{ V}}]
[V_b=2.5times 10^{-3}times 470 +0.7=boxed{1.9 textrm{ V}}]
Result
4 of 4
$$
V_b=1.9 textrm{ V}
$$
V_b=1.9 textrm{ V}
$$
Exercise 23
Step 1
1 of 4
In this problem, we are given a battery that powers two diodes connected in series with a given voltage in a circuit of known current and resistance. We should find the battery voltage.
Step 2
2 of 4
The equation that describes this process is Ohm’s law applied on the diode circuit
$$
V_b=IR+2V_d
$$
Step 3
3 of 4
center{Now, we can insert the given values to have that}
[V_b=2.5times 10^{-3}times 470 +2times 0.7=boxed{2.6 textrm{ V}}]
[V_b=2.5times 10^{-3}times 470 +2times 0.7=boxed{2.6 textrm{ V}}]
Result
4 of 4
$$
V_b=2.6 textrm{ V}
$$
V_b=2.6 textrm{ V}
$$
Exercise 24
Step 1
1 of 1
The positive end of the first diode when we follow the direction of the current (anode) should be attached to the cathode of the second diode where the anode of the second diode should be facing the positive voltage.
Exercise 26
Step 1
1 of 4
In this problem, we are given a battery that powers a diode with a given voltage in a circuit of known current, and resistance. We should find the battery voltage.
Step 2
2 of 4
The equation that describes this process is Ohm’s law applied on the diode circuit
$$
V_b=IR+V_d
$$
Step 3
3 of 4
center{Now, we can insert the given values to have that}
[V_b=12times 10^{-3}times 470 +0.4=boxed{6.04 textrm{ V}}]
[V_b=12times 10^{-3}times 470 +0.4=boxed{6.04 textrm{ V}}]
Result
4 of 4
$$
V_b=6.04 textrm{ V}
$$
V_b=6.04 textrm{ V}
$$
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