A 5 A source has a shunt internal resistance of 100Ω. The maximum power that can be delivered to a load isa)625 Wb)1250 Wc)325 Wd)none of aboveCorrect answer is option 'A'. Can you explain this answer?

Question

A 5 A source has a shunt internal resistance of 100&Omega;. The maximum power that can be delivered to a load isa)625 Wb)1250 Wc)325 Wd)none of aboveCorrect answer is option 'A'. Can you explain this answer?

Answer

Explanation:

Given Data:
- Shunt internal resistance, R = 100 Ω
- Maximum power delivered to the load, Pmax

Formula:
- The power delivered to the load is maximum when the load resistance is equal to the internal resistance of the source. This condition is known as Maximum Power Transfer Theorem.
- The formula for power delivered to the load is given by:
P = V² / (4R)

Calculation:
- Given R = 100 Ω
- To find the maximum power delivered to the load, we need to substitute R = 100 Ω in the formula for power.

Substitute the values:
Pmax = V² / (4 * 100)
Pmax = V² / 400

Now, we need to find the value of V:
- The maximum power is delivered when the load resistance is equal to the internal resistance of the source.
- Therefore, load resistance, RL = 100 Ω
- Total resistance, RT = R + RL = 100 + 100 = 200 Ω

Using Ohm's Law:
V = IR
V = I * 200

Substitute the value of V:
Pmax = (200I)² / 400
Pmax = 40000I² / 400
Pmax = 100I²

Maximum power delivered to the load is given by:
Pmax = 100I²
- Since power is maximum when load resistance is equal to source resistance, the above expression simplifies to:
Pmax = V² / (4R)
Pmax = 100

Therefore, the maximum power that can be delivered to the load is 625 W (Option A).

Answer

Here, I_N=5 A and R_N=100.
For maximum in the load,R_L=R_N=100Ω Therefore, current through load
=I_N/2=5/2=2.5A

Answer

Here, I_N=5 A and R_N=100.
For maximum in the load,R_L=R_N=100Ω Therefore, current through load
=I_N/2=5/2=2.5A

Answer

Here, I_N=5 A and R_N=100.
For maximum in the load,R_L=R_N=100Ω Therefore, current through load
=I_N/2=5/2=2.5A

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