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.

Result

I =  0.6 A
U2 =  11.88 V

Solution:

U1=12 V R1=0.2 Ω R2=19.8 Ω  I=U1R1+R2=120.2+19.8=35=0.6 AU_{1}=12 \ \text{V} \ \\ R_{1}=0.2 \ Ω \ \\ R_{2}=19.8 \ Ω \ \\ \ \\ I=\dfrac{ U_{1} }{ R_{1}+R_{2} }=\dfrac{ 12 }{ 0.2+19.8 }=\dfrac{ 3 }{ 5 }=0.6 \ \text{A}
U2=I⋅ R2=0.6⋅ 19.8=29725=11.88 VU_{2}=I \cdot \ R_{2}=0.6 \cdot \ 19.8=\dfrac{ 297 }{ 25 }=11.88 \ \text{V}

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