Source internal resistance

A load of 23.5 ohm is connected to an ideal voltage source of 30V, and a current of 1.2A passes through the circuit. Determine the magnitude of the source's internal resistance and the short-circuit current.

Final Answer:

R₂ =  1.5 Ω
I₂ =  20 A

Step-by-step explanation:

$$\begin{gathered} U=30\text{V} \\ {R}_1=23.5\Omega \\ I=1.2\text{A} \\ I=U/R \\ U=(R_1+R_2)\cdot I \\ {R}_2=U/I-R_1=30/1.2-23.5=1.5\Omega \end{gathered}$$

$I_2=\frac{U}{R_2}=\frac{30}{1.5}=20\text{A}$

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