Fall 2011_PHY301_4_SOL

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```					Assignment 4(Fall 2011)
(Solution)
Circuit Theory (PHY301)
Marks: 20
Due Date: Jan 24, 2012

Q.1
For the circuits shown in the Figure below using germanium diodes, find the values of the
voltages and currents indicated, mention the units of each derived value.

Sol.
Since we are using germanium diode, and voltage drop across the germanium diode is 0.3V.

(A) Diode On by inspection,
Diode   forward biased  V=5-0.3=4.7V
I = {4.7}/10 =-0.47mA
(B) Diode Off by inspection,
Diode   Open circuit  V=+5V
I = {5-(5)}/10 =0
(C) Diode On by inspection,
Diode   Forward bised V= 5V-0.3V=4.7V
I = {4.7}/10 =0.47mA
(D) Diode Off by inspection,
Diode   Open circuit  V=-5V
I = {5-(5)}/10 =0
Q.2:
For the given network, Calculate the voltage across the load, and current across the load.

Solution:
N1  20
N2  1
V1  100 A
I1  1A
Using turn ratio formula for voltage:
Putting the Values:
N 2 V2

N1 V1
1    V
 2
20 100
1
 100  V2
20
V2  5volts
Using turn ratio formula for Current:
Now
N 2 I1

N1 I 2
1   1

20 I 2
I 2  20 A
Q.3:
In order to get full input signal at output side, full wave rectifier OR full wave bridge
rectifier is used. Tell which one is more preferred and why?
Ans:
The bridge rectifier is more preferred over the full wave rectifier because
it does not require the use of center-tapped transformer, and therefore can be coupled
directly to the ac power line, if desired, therefore reducing size and cost.
By means of a transformer with the equivalent resultant electrical energy bring into being
a peak amount produced electrical energy that is practically two times the voltage of the
full brandish center-tapped rectifier. This consequence in the elevated dc electrical energy
from the contribute.

```
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Description: Past assignments, CS201, CS301, CS304, CS601, Pak301, Eng201, PHY301
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