Closed circuit

In a closed circuit, there is a voltage source with U1 = 12 V and with an internal resistance R1 = 0.2 Ω. The external resistance is R2 = 19.8 Ω.
Determine the electric current and terminal voltage.

Correct result:

I =  0.6 A
U₂ =  11.88 V

Solution:

$$\begin{gathered} U_1=12\text{ V } \\ {R}_1=0.2\mathrm{\Omega } \\ {R}_2=19.8\mathrm{\Omega } \\ I={\displaystyle \frac{U_1}{R_1+R_2}}={\displaystyle \frac{12}{0.2+19.8}}={\displaystyle \frac{3}{5}}={\displaystyle \frac{3}{5}}\text{ A}=0.6\text{ A} \end{gathered}$$

$U_2=I\cdot R_2=0.6\cdot 19.8={\displaystyle \frac{297}{25}}={\displaystyle \frac{297}{25}}\text{ V}=11.88\text{ V}$

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