To determine the power transfer efficiency of the voltage source when a 15Ω load is connected, we need to understand the behavior of the circuit, which includes a voltage source with internal resistance.
First, define the voltage across the source as V and the internal resistance as R. Two scenarios are given:
Scenario 1: When a 5Ω load is connected, the current I is 2A. The voltage equation is: V = 2(5 + R)
Scenario 2: With a 10Ω load, the current is 1.6A, giving the equation: V = 1.6(10 + R)
.
Now, for a 15Ω load, calculate the current:
Using the formula
I = V / (R + Load)
Substitute the values:
I = 40/(15 + 15) = 40/30 = 4/3 A
Using
2(5 + 15) = V, we find
V = 40 volts
Substituting R back into one of the original equations gives the voltage V:
By solving these equations simultaneously:
From 2(5 + R) = 1.6(10 + R), expand and simplify to obtain:
10 + 2R = 16 + 1.6R
Simplifying further: 0.4R = 6
Thus, R = 15Ω
Calculate the efficiency of power transfer to the load:
The power delivered to the load is: PL = (4/3)^2 * 15
The total power from the source: Ptotal = 40 * 4/3
Efficiency η is calculated as: η = (PL / Ptotal) * 100%
Thus, η = ((4/3 * 15) / 40) * 100% = 50%
The correct answer is 50%, indicating that with a 15Ω load, 50% of the power is efficiently transferred from the source to the load