Answered: For the circuit in the figure, find the…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

For the circuit in the figure, find the ones requested in the questions by using the following values?

VS=9 V, RA=6 kΩ, RB=3 kΩ, RC=4 kΩ

ID1=?

Expert Solution

Calculation:

Given data:

$V S = 9 V$

$R A = 6 k Ω$

$R B = 3 k Ω$

$R C = 4 k Ω$

A cut-in voltage of

$S_{i} = 0.7 V G e = 0.3 V$

Assume all diode's are open.

Then there is no path for the current to flow.

Consider $D_{1}$ is closed with a drop of $0.7 V$ .

Now,

$V_{D 2} = 0.7 V$

$V_{D 3} = 0.3 V$

Therefore,

$V_{D 3} < V_{D 2}$

Thus, $G e$ diode $D_{3}$ conducts first leaving $D_{2}$ in non-conducting state.

The result is as given below in circuit diagram.

Here, $D_{1}$ and $D_{3}$ are conducting and $D_{2}$ is cut-off.

Therefore,

$\begin{array}{rcl} I_{D 1} & = & \frac{V S - V_{D 2} - V_{D 3}}{(\frac{R A}{R B}) + R C} \\ = & \frac{9 V - 0.7 V - 0.3 V}{(\frac{6 k Ω}{3 k Ω}) + 4 k Ω} \\ = & \frac{8 V}{6 k Ω} \end{array}$

$\begin{array}{rcl} I_{D 1} & = & 1.333 m A \end{array}$

Here, only $D_{3}$ conducts and $D_{2}$ does not because $D_{3}$ cut-in voltage is $0.3 V$ after which it does not allow the voltage across $D_{2}$ to increases more than $0.3 V$ due to which $0.7 V$ is never reached across $D_{2}$ and it is cut off but $D_{1}$ and $D_{3}$ are conducting.

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