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Answer :
1) The Thevenin equivalent voltage for the circuit is [tex]V_{T} =12.5V[/tex]. 2. The Thevenin equivalent resistance for the circuit is R_{T}=0.15Ω. 3) The Norton equivalent current for the circuit is [tex]I_{N}=83.3A[/tex].
Refer to the figure below. Since open-circuit voltage is 12.5 V, then the voltage E=12.5 V also. This is value of Thevenin equivalent voltage VT.
Refer to the figure below. The current flowing through 0.1-Ω resistance is 50 A, then the Thevenin resistance is:
[tex]\frac{12.5}{R_{T}+0.1 }=50\\R_{T}=\frac{12.5}{50}-0.1\\R_{T}=0.15[/tex] Ω
Norton equivalent current is:
[tex]I_{N}=\frac{V_{T}}{R_{T}} =12.5/0.15 A=83.3[/tex]A
A short circuit can receive 484.6 A of current from the battery. Circuits that are equal to Thevenin and Norton are given below. Thevenin equivalent circuit appears more plausible because the amount of energy in the battery remains constant while it is open-circuit. No matter whether a load is connected or not in the Norton equivalent circuit, the battery will eventually deplete.
[tex]V_{T} =12.5V, R_{T}=0.15[/tex]Ω, [tex]I_{N}=83.3[/tex]A
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