Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is
- A0.8 Ω
- B1.4 Ω
- C2 Ω
- D2.8 Ω
Solution & Step-by-step Explanation
We obtain Thevenin equivalent of circuit B
Thevenin impedance:
Thevenin Voltage: Vₜₕ = 3∠0° V
Now, circuit becomes as
Current in the circuit,
I₁ = (10-3)/(2 + R)
Power transfer from circuit A to B
P = (I₁)²R + 3 I₁
= (42 + 70 R)/(2 + R)²
dP/dR = 0
⇒ (2 + R) [(2 + R) 70 – (42 + 70 R) 2] = 0
⇒ R = 0.8 Ω
Thevenin impedance:
Thevenin Voltage: Vₜₕ = 3∠0° V
Now, circuit becomes as
Current in the circuit,
I₁ = (10-3)/(2 + R)
Power transfer from circuit A to B
P = (I₁)²R + 3 I₁
= (42 + 70 R)/(2 + R)²
dP/dR = 0
⇒ (2 + R) [(2 + R) 70 – (42 + 70 R) 2] = 0
⇒ R = 0.8 Ω