To solve the problem, we use the properties of series circuits: current is equal through all components, and total voltage equals the sum of individual voltages.

Step 1: Calculate the voltage across (R_1)

Total voltage (U_{total}=12V), voltage across (R_2) (U_2=8V).
[U1 = U{total} - U_2 = 12V -8V =4V]

Step 2: Find (R_1)

Current through (R_1) (I_1=0.2A) (same as series current).
[R_1=\frac{U_1}{I_1}=\frac{4V}{0.2A}=20Ω]

Step3: Find (R_2)

Current through (R_2) (I_2=0.2A).
[R_2=\frac{U_2}{I_2}=\frac{8V}{0.2A}=40Ω]

Step4: Total resistance (R_{total})

[R_{total}=R_1+R_2=20Ω+40Ω=60Ω]

Answers:

1. (R_1=20Ω)
2. (R_2=40Ω)
3. (R_{total}=60Ω)

(\boxed{20}) (for (R_1)), (\boxed{40}) (for (R_2)), (\boxed{60}) (total resistance).

If only one answer is needed (assuming the question expects total resistance), then (\boxed{60}).

But based on the problem's implied questions:

• (R_1=\boxed{20Ω})
• (R_2=\boxed{40Ω})
• Total resistance=(\boxed{60Ω})

Final answer (total resistance): (\boxed{60})

(Note: If the question asks for all three, list them as above. But typically, such problems may prioritize total resistance, so (\boxed{60}) is the key result.)

(\boxed{60})

作者声明:本文包含人工智能生成内容。