Home Assignment
Numericals on Thevenin’s Theorem
Submission date: 4th April. Only Numericals: 1, 5, 6
1. Find the Thevenin’s resistance Thevenin’s equivalent voltage.
(Ans: Req = (5//10 + 3 = 6.33k ohm) and Veq = 20V)
2. Find the Thevenin’s resistance and Thevenin’s equivalent voltage.
This network contains a current source. So it will be important to you to understand how to
deal with a current source in thevenin’s theorem.
To find Req, short circuit voltage source, open circuit current source. For Veq, find voltage at
point C.
(Ans: Req = 7.1k ohm, Veq = 2.25V)
Nodal equation to point C
(VC – 0)/3 + (VC-VD)/ 2 = 0
5VC – 3VD = 0 --------------------------------(1)
Nodal equation to point D
(VD - VC)/2 + VD/10 + (VD -5V)/10 – 1 = 0
7VD – 5VC = 15 ---------------------------------------(2)
By solving 1&2
Vc = 2.25V = Veq
3. Find equivalent voltage and equivalent resistance.
4. Find equivalent voltage and equivalent resistance.
5. Use Thévenin's theorem to determine IL.
(Ans Veq = -2V, Req = 2ohm, IL = -2/5A)
6. Using Thevenin’s theorem, find the IL.
(Ans Veq = 8/7 V, Req = 10/7 ohm, IL = 8/31A)
7. Determine the unknown voltage V0 and the resistance R in the circuit. The maximum
power dissipation at the variable resistor RL is measured as 147mW when the
resistance RL is set to 3kΩ.
Ans. A) The maximum power dissipation at RL occurs at when the resistance is equal to
thevenin equivalent resistance Req (Maximum power transfer Theorem). Therefore here Req
= RL = 3kΩ.
Now find R.
(Ans. R = 4k Ω)
B) Now Find Veq. Use Maximum Power transfer Theorem.
Half of the power appears across the load and remaining half across the Req.
p = V2/R to RL
147mW = V2/ 3kΩ
The voltage across the load resistance for maximum power dissipation is 21V.
Therefore, the Veq = 21V (load) +21V (Req) = 42V.
Use this to find Vo.
C) The voltage Vo
Apply nodal equation to point C
(VC – V)/2kΩ + (VC – 0)/3kΩ – 12mA – 13mA = 0
(42 – V) /2kΩ + 42/3kΩ -25mA = 0
(42 – V) x 10-3/2 = 11x10-3
42 – V = 22
V = 20v
Applying nodal equation to point B
(0 – V1)/ 4kΩ + (0 - VC)/ 3kΩ + 13mA = 0
-V1/4 + (-42)/3 +13 = 0
-V1/4 -1 = 0
V1 = -4v
V0 + V1 = 20
Therefore Vo = 24 V