In the circuit shown below, determine the current in the circuit. How much power is dissipated by each resistor in the figure? R₁ = 120 9.0 V R₂= = 15 Ω wwwwww

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**Determine Current and Power in Series Circuit**

**Problem Overview:**
In the circuit shown below, determine the current in the circuit. How much power is dissipated by each resistor in the figure?

**Circuit Diagram:**
The circuit consists of a series connection with the following components:
- A voltage source of 9.0 volts (V).
- Two resistors:
  - \( R_1 \) with a resistance of 12 ohms (Ω).
  - \( R_2 \) with a resistance of 15 ohms (Ω).

**Analysis:**
1. **Total Resistance in the Circuit:**
   Since the resistors are in series, the total resistance \( R_t \) is the sum of the resistances:
   \[
   R_t = R_1 + R_2 = 12 \, \Omega + 15 \, \Omega = 27 \, \Omega
   \]

2. **Current in the Circuit:**
   Using Ohm's Law to determine the current \( I \) in the circuit:
   \[
   I = \frac{V}{R_t} = \frac{9.0 \, \text{V}}{27 \, \Omega} = 0.33 \, \text{A} \, (\text{or } \frac{1}{3} \, \text{A})
   \]

3. **Power Dissipated by Each Resistor:**
   - For \( R_1 \):
     \[
     P_1 = I^2 \times R_1 = (0.33 \, \text{A})^2 \times 12 \, \Omega = 0.11 \, \text{W}
     \]
     
   - For \( R_2 \):
     \[
     P_2 = I^2 \times R_2 = (0.33 \, \text{A})^2 \times 15 \, \Omega = 0.165 \, \text{W}
     \]

**Conclusion:**
The current in the circuit is 0.33 A. The power dissipated by resistor \( R_1 \) is 0.11 W, and the power dissipated by resistor \( R_2 \) is 0.165 W.
Transcribed Image Text:**Determine Current and Power in Series Circuit** **Problem Overview:** In the circuit shown below, determine the current in the circuit. How much power is dissipated by each resistor in the figure? **Circuit Diagram:** The circuit consists of a series connection with the following components: - A voltage source of 9.0 volts (V). - Two resistors: - \( R_1 \) with a resistance of 12 ohms (Ω). - \( R_2 \) with a resistance of 15 ohms (Ω). **Analysis:** 1. **Total Resistance in the Circuit:** Since the resistors are in series, the total resistance \( R_t \) is the sum of the resistances: \[ R_t = R_1 + R_2 = 12 \, \Omega + 15 \, \Omega = 27 \, \Omega \] 2. **Current in the Circuit:** Using Ohm's Law to determine the current \( I \) in the circuit: \[ I = \frac{V}{R_t} = \frac{9.0 \, \text{V}}{27 \, \Omega} = 0.33 \, \text{A} \, (\text{or } \frac{1}{3} \, \text{A}) \] 3. **Power Dissipated by Each Resistor:** - For \( R_1 \): \[ P_1 = I^2 \times R_1 = (0.33 \, \text{A})^2 \times 12 \, \Omega = 0.11 \, \text{W} \] - For \( R_2 \): \[ P_2 = I^2 \times R_2 = (0.33 \, \text{A})^2 \times 15 \, \Omega = 0.165 \, \text{W} \] **Conclusion:** The current in the circuit is 0.33 A. The power dissipated by resistor \( R_1 \) is 0.11 W, and the power dissipated by resistor \( R_2 \) is 0.165 W.
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