Electrical Engineering - Circuit Theorems and Conversions - Discussion
Discussion Forum : Circuit Theorems and Conversions - General Questions (Q.No. 1)
1.
Find the Thevenin equivalent (VTH and RTH) between terminals A and B of the circuit given below.
4.16 V, 120 
41.6 V, 120
4.16 V, 70
41.67 V, 70
Discussion:
43 comments Page 1 of 5.
Ravi said:
6 years ago
By Voltage dider rule(vth) = 100 * 120/(60 + 100 + 120) = 41.66.
And rth = (100 + 68)/(100 + 68 + 120) = 70.
And rth = (100 + 68)/(100 + 68 + 120) = 70.
(13)
Muzaffar said:
9 years ago
I = 100/288 = 0.347A.
Rth = 120/(100+68) = 70.
Vth= 0.347 * 120 = 41.6 v.
Rth = 120/(100+68) = 70.
Vth= 0.347 * 120 = 41.6 v.
(12)
Nagesh said:
1 decade ago
The resistances 68 and 100 are in series ; 68 + 100 = 168.
168 ohms is in parallel with 120 ohms. So,
Rth = (168 * 120) / (168 + 120) = 70 ohms.
Vth = R3 * (Vs/R1 + R2 + R3).
Vth = 41.67V.
168 ohms is in parallel with 120 ohms. So,
Rth = (168 * 120) / (168 + 120) = 70 ohms.
Vth = R3 * (Vs/R1 + R2 + R3).
Vth = 41.67V.
(11)
Balram jadhav said:
10 years ago
By using vtg division rule (Vth) = Vs * R3/(R1 + R2 + R3).
and Rth = R1 ll (R2 + R3).
and Rth = R1 ll (R2 + R3).
(7)
Swarnab Kanrar said:
2 years ago
Vth= Vs * R3/(R1+ R2 +R3)
= 100 * 120/(68+100+120)
= 41.66 V.
Rth = R3 * (R1 +R2)/(R1+ R2 +R3),
= 120 * (68+100)/(68+100+120),
= 70 ohm.
= 100 * 120/(68+100+120)
= 41.66 V.
Rth = R3 * (R1 +R2)/(R1+ R2 +R3),
= 120 * (68+100)/(68+100+120),
= 70 ohm.
(6)
Siddhartha said:
9 years ago
What if there is no resistance between A and B terminals?
(5)
Pinvikesh said:
2 decades ago
I am not getting answer for it and I also want the simplest way to do these problems of each modle to save time please send me and also your site is useful for alot us
(2)
Ayushi said:
1 decade ago
Actually your doubt is common one, which usually strikes the mind anirban but R3 is not the load resistance infact the load is on a b terminal, and the load is not shown at the terminals. So the answer given that is 70 is correct.
(1)
Prafull gadekar said:
1 decade ago
Total Resistance,
Rt=R1+R2+R3=120+68+100 = 288ohm.
I = V/R = 100/288 = 0.347A(flowing same current in all resistance).
Now, we want to find voltage across R3=120 ohm.
V = 120*0.347 = 41.67V
Now, Rth = R3//(R1+R2) = 70 ohm.
Rt=R1+R2+R3=120+68+100 = 288ohm.
I = V/R = 100/288 = 0.347A(flowing same current in all resistance).
Now, we want to find voltage across R3=120 ohm.
V = 120*0.347 = 41.67V
Now, Rth = R3//(R1+R2) = 70 ohm.
(1)
Keerthana said:
1 decade ago
1. Find Open Circuit Voltage(Vth):
Req = 68+100+120 = 288v.
Ieq = V/Req = 100/288 = .34722 A.
Voc =100-(68*.344722)-(100*.34722).
We get Vth = Voc = 41.667v.
Isc = 100/(68+100)= 0.5952A.
Rth = Vth/Isc.
Rth = 41.667/.5952 = 70 ohm.
Req = 68+100+120 = 288v.
Ieq = V/Req = 100/288 = .34722 A.
Voc =100-(68*.344722)-(100*.34722).
We get Vth = Voc = 41.667v.
Isc = 100/(68+100)= 0.5952A.
Rth = Vth/Isc.
Rth = 41.667/.5952 = 70 ohm.
(1)
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