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The given circuit shows a combination of resistors with values:

- \( E_T = 120 \, \text{V} \) (total voltage),
- \( I_T = 200 \, \text{mA} = 0.2 \, \text{A} \) (total current),
- \( R_1 = 1 \, \text{k}\Omega = 1000 \, \Omega \),
- \( R_3 = 5.6 \, \text{k}\Omega = 5600 \, \Omega \),
- \( R_2 = ? \) (unknown).

### Step 1: Determine the total resistance of the circuit
Using Ohm's law:

\[
R_T = \frac{E_T}{I_T} = \frac{120 \, \text{V}}{0.2 \, \text{A}} = 600 \, \Omega
\]

### Step 2: Understand the circuit configuration
The circuit shows \( R_1 \) in series with a parallel combination of \( R_2 \) and \( R_3 \).

- Total resistance, \( R_T \), is the sum of the series resistance \( R_1 \) and the equivalent resistance of \( R_2 \) and \( R_3 \) in parallel (\( R_{23} \)):

\[
R_T = R_1 + R_{23}
\]

### Step 3: Solve for \( R_{23} \)
\[
R_{23} = R_T - R_1 = 600 \, \Omega - 1000 \, \Omega = -400 \, \Omega
\]

This negative result indicates there was an error in interpretation

Solve with solution and explain the circuit with E_T = 120 V, I_T = 200 mA, R1 = 1 kΩ, R2 = ?, R3 = 5.6 kΩ.

Learn how to solve for R2 in a series-parallel circuit where the total voltage is 120V, the total current is 200mA, and the resistors R1 = 1kΩ, and R3 = 5.6kΩ. This problem involves Ohm's Law and the concept of parallel and series resistances.

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