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Assignment 6 (Spring 2010)

Circuit Theory (PHY301)

Marks: 20

Due Date: July 29, 2010

 

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Q.1

   A bridge rectifier is fed by a 25 Vac transformer. Determine the dc load voltage and        current for the circuit when it has a 2.5kΩload.

 

Q.2

If 10w of power is applied to primary of an ideal transformer with a turn ratio of 5, Find power delivered to the secondary load.

 

Q.3

   Find the value of R in series of silicon identical diode at drop of 2mA, if V0 = 4V when    IL =0, and V0 changes by 45mV per 2mA of load current. (Assume n =1)

 

        

 

 

 

 

 

 

 

                   

 

 

 

 

                                 ………….Good Luck……….

 

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Replies to This Discussion

Q.1
A bridge rectifier is fed by a 25 Vac transformer. Determine the dc load voltage and current for the circuit when it has a 2.5kΩload.

Sol.
With the 22Vac rated transformer, the peak secondary voltage is found as
V2(pk) = 25/0.707
=35.36 Vpk
The peak load voltage is now found as
VL(pk) = V2 – 1.4
VL(pk) =33.96Vpk
The dc load voltage is found as
Vave = 2VL(pk)/Π
=67.92/Π
Vave =21.61 Vdc
Finally the dc load current is found as
Iave = Vave/RL
= 21.61/2.5k
Iave = 8.64μA



Q.2
If 10w of power is applied to primary of an ideal transformer with a turn ratio of 5, Find power delivered to the secondary load.

Q.3
Find the value of R in series of silicon identical diode at drop of 2mA, if V0 = 4V when IL =0, and V0 changes by 45mV per 2mA of load current. (Assume n =1)












SOLUTION:
V0 = 4V, when IL =0, therefore each diode should exhibit a drop of 0.75V. If IL =0mA, then Vo changes by 45mV and a change due to each diode is 12mV.
Hence
rd = 12mV/2mA
=6 Ohms

but
rd = nVT/ID
6 = 1 x 25m/ID
ID = 4.1mA
Hence
15 – 3 – IDRD = 0
R = (15 – 3)/ID
= (15 – 3)/4.1m
= 2.92 k Ohms.
See the attached file for Assignment Solution
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