Question

1. The circuit shown uses a 9.0V zener diode. If the load resistance RL is equal to 1.5 k2, and the DC source
equals 24v, find the maximum value of resistor Rs required to maintain a constant voltage of 9V across the
load.
[Ans:Rs =2.5 K2 ]​

Answer 1

Answer:

At V  

B =9V

L = 4

​6×10 −3(-3 is the power of 10,every where)

=1.5×10−3 A

i = R

​8−6×10 −3

=2×10 −3A

∴i ze.

R =i  

load −i  

=0.5×10 −3 A

​At V  

​B=16V  

i =L

=1.5×10 − 3A

​i =R

(16−6)×10 −3

=10×10 −3A

∴i ze.

R =i  

L −i =8.5×10  −3

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Explanation:

if you put your values in this you find your ans.

hope you

understood it

Answer 2

Answer:

The maximum value of the resistor  $R_s$   required to maintain a constant voltage of 9V across the load is 2.5kΩ.

Explanation:

• The resistors $R_s$  and  $R_L$ are in series.
• The same amount of current will flow through $R_s$  and  $R_L$ .

So, the current flowing through  $R_L$ will be given as,

$I=\frac{V_z}{R_L}$      (1)

Where,

I=current through  $R_L$

$V_z$=potential difference across the Zener diode

$R_L$=load resistance

From the question we have,

$V_z$=9 volt

$R_L$=1.5kΩ=1.5×10³Ω

The voltage supplied by DC source=24 Volt

By substituting the values in equation (1) we get;

$I=\frac{9}{1.5\times 10^3}$

$I=6\times 10^-^3A$    (2)

The maximum value of the resistor  $R_s$ will be given as,

$R_s=\frac{V-V_z}{I}$      (3)

By substituting all the required values in equation (3) we get;

$R_s=\frac{24-9}{6\times 10^-^3}$

$R_s=\frac{15}{6\times 10^-^3}$

$R_s=2.5\times 10^-^3$

$R_s=2.5k\Omega$

Hence, the maximum value of the resistor  $R_s$   required to maintain a constant voltage of 9V across the load is 2.5kΩ.

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