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Thevenin: 100, 60, 150 ohm circuit

Short answer

Thevenin's theorem: replace network with Vth and Rth. Open-circuit voltage Vth = (100−60)/(100+60) gives 75 V; Rth = 100∥60 = 37.5 Ω. Current through 150 Ω is I = 75/(37.5+150) = 0.4 A.

Find the current I in the 150 Ω resistor R3 using Thevenin's theorem. [Figure: E1 = 100 V with R1 = 100 Ω in series in the left branch; E2 = 60 V in the middle branch (with R2 = 60 Ω in series between the middle branch and the load branch); R3 = 150 Ω load carrying current I.]

Thevenin's theorem: a 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 and Rth is the resistance seen from the terminals with all independent sources set to zero.

Assumption: from the description, E1 = 100 V with R1 = 100 Ω, and E2 = 60 V with R2 = 60 Ω are two parallel source branches across the nodes to which the 150 Ω load R3 is connected. Both batteries are taken with their positive ends toward the top node.

Step 1: remove R3, find Vth

Loop current with R3 open:

I' = (E1 − E2)/(R1 + R2) = (100 − 60)/(100 + 60) = 0.25 A

Vth = E1 − I'R1 = 100 − 0.25 × 100 = 75 V

(Check: E2 + I'R2 = 60 + 0.25 × 60 = 75 V.)

Step 2: find Rth

Short both sources: Rth = R1 ∥ R2 = (100 × 60)/(100 + 60) = 37.5 Ω

Step 3: load current

 Rth = 37.5 ohm
 +--/\/\/\--+
 |            |
 Vth = 75 V   R3 = 150 ohm  ->  I
 |            |
 +------------+

I = Vth/(Rth + R3) = 75/(37.5 + 150) = 75/187.5

Answer: I = 0.4 A through the 150 Ω resistor

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