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Thevenin current through RL and maximum power

Short answer

Using Thevenin’s theorem, Vth = 11.5 V, Rth = 1.5 Ω. With RL = 2 kΩ, the load current is IL = Vth/(Rth+RL) ≈ 5.75 mA. Maximum power transfer requires RL = Rth = 1.5 Ω, and the maximum power is Pmax = Vth²/(4Rth) = 22.04 W.

In a given network, find the current flowing through the load resistance RL by using Thevenin's theorem. What should be the value of RL for maximum power to transfer? Also find the maximum power. [Figure: a 12 V battery (positive terminal at top) on the left, connected across a 4 Ω resistor (top node P to bottom line); a 2 Ω resistor from node P to node Q; from node Q to the bottom line a branch of 6 Ω in series with a 10 V battery (positive terminal toward the 6 Ω, i.e. upward); load RL = 2 kΩ connected from node Q to the bottom line.]

Thevenin's theorem: any linear two-terminal network can be replaced by a voltage source Vth in series with a resistance Rth, where Vth is the open-circuit voltage at the terminals and Rth is the resistance seen from the terminals with independent sources replaced by their internal resistances.

Take the bottom line as reference. Node P is fixed at 12 V by the battery (ideal), so the 4 Ω across the battery does not affect the rest of the circuit.

 P   2Ω    Q
 o--/\/\---o----+-----+
 |         |    |     |
12V  4Ω   6Ω    RL    (open for Vth)
 |         |    |
 |       +10V   |
 o---------o----+

Step 1: Thevenin voltage (RL removed)

Current flows from P (12 V) through 2 Ω and 6 Ω into the + terminal of the 10 V battery:

I   = (12 − 10)/(2 + 6) = 2/8 = 0.25 A
Vth = VQ = 12 − I × 2  = 12 − 0.5 = 11.5 V
check: VQ = 10 + 0.25 × 6 = 11.5 V

Step 2: Thevenin resistance

Short both batteries. The 4 Ω is shorted by the 12 V source, so from Q we see 2 Ω in parallel with 6 Ω:

Rth = (2 × 6)/(2 + 6) = 1.5 Ω

Step 3: Load current (RL = 2 kΩ)

IL = Vth/(Rth + RL)
   = 11.5/(1.5 + 2000)
   = 5.746 × 10⁻³ A

Answer: IL ≈ 5.75 mA (power in the 2 kΩ load ≈ 66 mW).

Step 4: Maximum power transfer

Maximum power is delivered when RL = Rth:

RL  = Rth = 1.5 Ω
Pmax = Vth²/(4 Rth) = (11.5)²/(4 × 1.5)
     = 132.25/6 = 22.04 W

Answer: RL = 1.5 Ω for maximum power; Pmax = 22.04 W.

The equivalent circuit is 11.5 V in series with 1.5 Ω feeding RL. With RL = 2 kΩ the load is badly mismatched, so it takes only about 0.3% of the maximum possible power.

✓ Answer checked by Claude Opus 5.5 (high effort)

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