Test: Network Theorems - 1 - Electrical Engineering (EE) MCQ

After killing all source equivalent resistance is R Open circuit voltage = v₁

The short circuit current across the terminal is

For the calculation of R_TH if we kill the sources then 20 Ω resistance is inactive because 5 A source will be open circuit

After killing all source,

If we solve this circuit direct, we have to deal with three variable. But by simple manipulation variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent

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)

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Now, for a 15Ω load, calculate the current:

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: P_L = (4/3)^2 * 15
The total power from the source: P_total = 40 * 4/3
Efficiency η is calculated as: η = (P_L / P_total) * 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