Respective 25361

Three resistors are connected according to the scheme (R1 in parallel with R2 and series with R3: R1||R2 + R3) so that R1=0.6 Ω, R2=2 Ω, R3=3 Ω, current I1 = 10 A. Calculate all the currents passing through the respective resistors R1, R2, R3 and all the voltages on the respective resistors U1, U2, U3.

Correct answer:

I₂ =  3 A
I₃ =  13 A
U₁ =  6 V
U₂ =  6 V
U₃ =  39 V

Step-by-step explanation:

$$\begin{gathered} R_1=0.6\Omega \\ {R}_2=2\Omega \\ {R}_3=3\Omega \\ {I}_1=10\text{A} \\ {U}_1=R_1\cdot I_1=0.6\cdot 10=6\text{V} \\ {U}_2=U_1=6\text{V} \\ {I}_2=U_2/R_2=6/2=3\text{A} \end{gathered}$$

$I_3=I_1+I_2=10+3=13\text{A}$

$U_3=I_3\cdot R_3=13\cdot 3=39\text{V}$

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