5. For the circuit shown find the (a) voltage across the 6.0 Q resistor and (b) the power supplied by the battery. 8 = 12.0 V R = 3.0 Q R = 2.0 Q R = 6.0 Q R = 1.5 Q R = 3.0 Q

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For the circuit shown, find the (a) voltage across the 6.0 ohm resistor and (b) the power supplied by the battery. 

**Text Transcription:**

5. For the circuit shown, find the (a) voltage across the 6.0 Ω resistor and (b) the power supplied by the battery.

**Diagram Description:**

The circuit diagram represents an electrical circuit with one voltage source and several resistors:

- The battery has an electromotive force (ε) of 12.0 volts.
- There are five resistors in the circuit with the following resistances:
  - A 1.5 Ω resistor is in series with the battery.
  - A 3.0 Ω resistor follows in series.
  - After the 3.0 Ω resistor, the circuit branches into two parallel paths:
    - One branch has a 2.0 Ω resistor.
    - The other branch has a 6.0 Ω resistor.
  - These two branches reconverge and connect to another 3.0 Ω resistor in series.

To solve the problem, identify the equivalent resistance of the circuit, use Ohm's Law (V = IR) to find the current, and apply the proper equations for voltage in series and parallel circuits. Then calculate the power supplied by the battery using P = IV.
Transcribed Image Text:**Text Transcription:** 5. For the circuit shown, find the (a) voltage across the 6.0 Ω resistor and (b) the power supplied by the battery. **Diagram Description:** The circuit diagram represents an electrical circuit with one voltage source and several resistors: - The battery has an electromotive force (ε) of 12.0 volts. - There are five resistors in the circuit with the following resistances: - A 1.5 Ω resistor is in series with the battery. - A 3.0 Ω resistor follows in series. - After the 3.0 Ω resistor, the circuit branches into two parallel paths: - One branch has a 2.0 Ω resistor. - The other branch has a 6.0 Ω resistor. - These two branches reconverge and connect to another 3.0 Ω resistor in series. To solve the problem, identify the equivalent resistance of the circuit, use Ohm's Law (V = IR) to find the current, and apply the proper equations for voltage in series and parallel circuits. Then calculate the power supplied by the battery using P = IV.
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