Short-circuit and fuse

An external circuit with a resistance of 3.8 Ω is connected to an electrical voltage source with Ue=12 V. A current of 3 A passes through the circuit.
Find:
1. source terminal voltage,
2. the internal resistance of the source,
3. maximum short-circuit current.

Correct answer:

U₂ =  11.4 V
R₂ =  0.2 Ω
I₂ =  60 A

Step-by-step explanation:

$$\begin{gathered} R=3.8\Omega \\ I=3\text{A} \\ {U}_1=12\text{V} \\ {U}_2=R\cdot I=3.8\cdot 3\text{V}=\frac{57}{5}\text{V}=11.4\text{V} \end{gathered}$$

$R_2=\frac{U_1-U_2}{I}=\frac{12-11.4}{3}=\frac{1}{5}\Omega =0.2\Omega$

$I_2=\frac{U_1}{R_2}=\frac{12}{\frac{1}{5}}=60\text{A}$

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