Find the Thevenin equivalent circuit looking into terminals a and b in the circu
ID: 1715770 • Letter: F
Question
Find the Thevenin equivalent circuit looking into terminals a and b in the circuit below by finding V_infinity and infinity. Find the Norton equivalent circuit for circuit below. Consider the circuits given in problems 1 and 2 above. Can we find R_v for circuit given in problem 1 by zeroing the sources? Why or why not? If yes. find R^ using this technique. Can we find Rr. by zeroing sources for problem 2? Why or why not? If yes. find R_0 using this technique. Find the voltage v, in the circuit below using mesh current analysis.Explanation / Answer
1) HERE CURRENT SOURCE i.e 2 amps is converting to 300 volts voltage source.
apply nodal analysis at node v1
(v1-300)/225+V1/20+(V1-V2-75)/75=0
V1(1/225+1/20+1/75)+V2(-1/75)=1+300/225
apply nodal analysis at node V2
V2/100+(V2+75-V1)/75=0
V1(-1/75)+V2(1/100+1/75)=-1
FROM THE ABOVE EQUATIONS
V1=29.28 VOLTS
V2=-26.72 VOLTS
OPEN CIRCUIT VOLTAGE VOC=-26.72 VOLTS
HERE THERE IS A SHORT CIRCUIT ACROSS NODES a AND b.
so 100 ohms resistor is neglected. because there is a short circuit across this resistor.
consider left side loop current as i1, second loop current as i2, third loop current as i3 and top loop current is i4.
all the loop currents are clockwise direction.
from the circuit i1= 2 amps.
apply kvl to second loop
150(i2-2)+75i2+20(i2-i3)=0
245i2-20i3=300
from the circuit i4-i3=1 amps
apply kvl to loops 3 and 4.
75i4+20(i3-i2)=0
-20i2+20i3+75i4=0
from the above equations
i2=1.18 amps
i3=-0.54 amps
i4=0.459 amps
short circuit current isc=0.459 amps.
2) open circuit voltage across nodes a and b.
consider left side node as v1 and right side node is v2.
apply nodal analysis at node v1.
(v1-10)/5+0.5ix+(v1-v2)/10=0
from the circuit ix=(10-v1)/5
(v1-10)/5+0.5*(10-v1)/5+(v1-v2)/10=0
v1(1/5-0.5/5+1/10)+v2(-1/10)=(10/5)-1
apply nodal analysis at node v2.
v2/10+(v2-v1)/10=0
v1(-1/10)+v2(1/5)=0
from the above equations.
v1=6.66 volts
v2=3.33 volts
voc=3.33 volts.
short circuit current isc across a and b:
now 10 ohms resistor is connected across short circuit branch.so 10 ohms resistor is neglected.
consider node voltage is v.
apply nodal analysis at node v.
(v-10)/5+v/10+0.5ix=0
from the circuit ix=(10-v)/5.
(v-10)/5+v/10+0.5*(10-v)/5=0
v(1/5+1/10-1/10)=2-1
v= 5 volts
short circuit current isc=v/10=5/10=0.5 amps
4) consider left side loop current is i1.
top loop current is i2
right side loop current is i3.
all the loop currents are clockwise direction.
from the circuit i1=1 amps
apply kvl to loop 2
i2+1(i2-i3)+2(i2-i1)=0
4i2-i3=2
apply kvl to loop 3.
2i3-2vx+1(i3-i2)=0
2i3-2*2i3+1(i3-i2)=0
-i2-i3=0
from the above equations
i2=0.4 amps
i3=-0.4 amps
vx=2i3=2*-0.4=-0.8 volts.
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