The current in the figure below is 0.59 A. (Let the internal resistance of battery 1 be r₁ = 1.52 and the internal resistance of battery 2 be r₂ = 1.2 2.) n 6.0 V 9.0 V 12 ww WH Battery 1 Battery 2 a b (a) Find the voltage drop Vab. (b) Find the power dissipated in R. (c) Find the useful electric power supplied by battery 2. (Note: Do not forget the internal resistance of battery 2.) (d) Find the rate at which chemical energy is stored in battery 1. (Note: Do not include power dissipated in the internal resistor of battery 1 in your answer.) R

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### Electrical Circuit Analysis - Example Problem

The diagram depicts an electrical circuit with two batteries and various resistors. Here are the details and tasks for the circuit analysis:

**Given Information:**
- **Current (I):** 0.59 A
- **Battery 1:**
  - EMF: 6.0 V
  - Internal Resistance (\(r_1\)): 1.5 Ω
- **Battery 2:**
  - EMF: 9.0 V
  - Internal Resistance (\(r_2\)): 1.2 Ω
- **Resistor (R):** Connects points \( a \) and \( b \)

**Tasks:**

#### (a) Find the voltage drop \( V_{ab} \).
\[ V_{ab} = \]

#### (b) Find the power dissipated in \( R \).
\[ P_R = \]

#### (c) Find the useful electric power supplied by battery 2. *(Note: Do not forget the internal resistance of battery 2.)*
\[ P_{useful, battery 2} = \]

#### (d) Find the rate at which chemical energy is stored in battery 1. *(Note: Do not include power dissipated in the internal resistor of battery 1 in your answer.)*
\[ P_{storage, battery 1} = \]

### Explanation of the Circuit Diagram:
- **Components:**
  - **Battery 1** (EMF = 6.0 V, Internal Resistance = 1.5 Ω)
  - **Battery 2** (EMF = 9.0 V, Internal Resistance = 1.2 Ω)
  - **Resistor \( R \)** (Connecting points \( a \) and \( b \))
- **Current Flow:** The circuit shows that a current of 0.59 A flows through each component.

**Circuit Path:**
1. From Battery 1, the current passes through its internal resistance (\( r_1 \)), delivering a part of its voltage to another resistance (\( R \)).
2. The voltage drop across \( R \) is determined by the potential difference between points \( a \) and \( b \).
3. The circuit continues with Battery 2, considering its EMF and internal resistance (\( r_2 \)).

### Calculation Steps:
**(a) Voltage drop \( V_{ab} \):**
Transcribed Image Text:### Electrical Circuit Analysis - Example Problem The diagram depicts an electrical circuit with two batteries and various resistors. Here are the details and tasks for the circuit analysis: **Given Information:** - **Current (I):** 0.59 A - **Battery 1:** - EMF: 6.0 V - Internal Resistance (\(r_1\)): 1.5 Ω - **Battery 2:** - EMF: 9.0 V - Internal Resistance (\(r_2\)): 1.2 Ω - **Resistor (R):** Connects points \( a \) and \( b \) **Tasks:** #### (a) Find the voltage drop \( V_{ab} \). \[ V_{ab} = \] #### (b) Find the power dissipated in \( R \). \[ P_R = \] #### (c) Find the useful electric power supplied by battery 2. *(Note: Do not forget the internal resistance of battery 2.)* \[ P_{useful, battery 2} = \] #### (d) Find the rate at which chemical energy is stored in battery 1. *(Note: Do not include power dissipated in the internal resistor of battery 1 in your answer.)* \[ P_{storage, battery 1} = \] ### Explanation of the Circuit Diagram: - **Components:** - **Battery 1** (EMF = 6.0 V, Internal Resistance = 1.5 Ω) - **Battery 2** (EMF = 9.0 V, Internal Resistance = 1.2 Ω) - **Resistor \( R \)** (Connecting points \( a \) and \( b \)) - **Current Flow:** The circuit shows that a current of 0.59 A flows through each component. **Circuit Path:** 1. From Battery 1, the current passes through its internal resistance (\( r_1 \)), delivering a part of its voltage to another resistance (\( R \)). 2. The voltage drop across \( R \) is determined by the potential difference between points \( a \) and \( b \). 3. The circuit continues with Battery 2, considering its EMF and internal resistance (\( r_2 \)). ### Calculation Steps: **(a) Voltage drop \( V_{ab} \):**
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