R1 TE R3 ww R2 The values of the resistors in this circuit are: R1 = 5 ohms R2 = 3 ohms R3 = 7 ohms The power consumption of R2 is 12 W. Calculate the current through R3.

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**Electrical Circuit Analysis: Calculating Current through Resistors**

**Description of the Circuit:**
The image depicts an electrical circuit with three resistors, R1, R2, and R3. These resistors are connected in a combination of series and parallel configurations with a power source.

**Configuration:**
- Resistor R1 (5 ohms) and Resistor R2 (3 ohms) are connected in series with each other.
- Resistor R3 (7 ohms) is connected in parallel with the series combination of R1 and R2.

**Given Values:**
- R1 = 5 ohms
- R2 = 3 ohms
- R3 = 7 ohms

**Additional Information:**
- Power consumption of R2 = 12 W

**Objective:**
Calculate the current through Resistor R3.

**Step-by-Step Solution:**

1. **Determine the voltage across R2:**
   The power consumption formula is \( P = I^2 \cdot R \).
   Given that \( P = 12 \,W \) and \( R = 3 \, \Omega \) (for R2),
   \[
   12 = I_{R2}^2 \cdot 3
   \]
   Solving for \( I_{R2} \):
   \[
   I_{R2}^2 = \frac{12}{3} = 4
   \]
   \[
   I_{R2} = \sqrt{4} = 2 \,A
   \]
   Thus, the current through R2 is 2 A.

2. **Calculate the voltage across R2 (V_R2):**
   Using Ohm's Law \( V = I \cdot R \):
   \[
   V_{R2} = I_{R2} \cdot R2 = 2 \, A \cdot 3 \, \Omega = 6 \, V
   \]

3. **Determine the voltage across R3 (V_R3):**
   Since R3 is in parallel with the series combination of R1 and R2, the voltage across R3 is the same as the voltage across the series combination (which is 6 V).

4. **Calculate the current through R3 (I_R3):**
   Again, using Ohm's Law \( I = \frac
Transcribed Image Text:**Electrical Circuit Analysis: Calculating Current through Resistors** **Description of the Circuit:** The image depicts an electrical circuit with three resistors, R1, R2, and R3. These resistors are connected in a combination of series and parallel configurations with a power source. **Configuration:** - Resistor R1 (5 ohms) and Resistor R2 (3 ohms) are connected in series with each other. - Resistor R3 (7 ohms) is connected in parallel with the series combination of R1 and R2. **Given Values:** - R1 = 5 ohms - R2 = 3 ohms - R3 = 7 ohms **Additional Information:** - Power consumption of R2 = 12 W **Objective:** Calculate the current through Resistor R3. **Step-by-Step Solution:** 1. **Determine the voltage across R2:** The power consumption formula is \( P = I^2 \cdot R \). Given that \( P = 12 \,W \) and \( R = 3 \, \Omega \) (for R2), \[ 12 = I_{R2}^2 \cdot 3 \] Solving for \( I_{R2} \): \[ I_{R2}^2 = \frac{12}{3} = 4 \] \[ I_{R2} = \sqrt{4} = 2 \,A \] Thus, the current through R2 is 2 A. 2. **Calculate the voltage across R2 (V_R2):** Using Ohm's Law \( V = I \cdot R \): \[ V_{R2} = I_{R2} \cdot R2 = 2 \, A \cdot 3 \, \Omega = 6 \, V \] 3. **Determine the voltage across R3 (V_R3):** Since R3 is in parallel with the series combination of R1 and R2, the voltage across R3 is the same as the voltage across the series combination (which is 6 V). 4. **Calculate the current through R3 (I_R3):** Again, using Ohm's Law \( I = \frac
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