60 V, 2 0 20Ω 30 0 in 25 0 A D

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a) Calculate the equivalent resistance of the three resistors connected in series and calculate the equivalent resistance for the entire circuit.

**Understanding Potential Difference in a Series Circuit**

**Objective:**
To calculate the potential difference \( V_A - V_B \) and \( V_C - V_D \) in the given circuit. Note that the battery has an internal resistance.

**Circuit Description:**

The circuit comprises the following components:
1. A battery with an electromotive force (emf) of \( 60 \text{ V} \) and an internal resistance of \( 2 \Omega \).
2. Three resistors connected in series:
   - \( 20 \Omega \) resistor between points A and B,
   - \( 30 \Omega \) resistor between points B and C,
   - \( 25 \Omega \) resistor between points C and D.

**Detailed Description:**

- The series circuit begins with the positive terminal of the battery.
- The negative terminal of the battery is connected through an internal resistance of \( 2 \Omega \).
- Point A to B consists of a \( 20 \Omega \) resistor.
- Point B to C consists of a \( 30 \Omega \) resistor.
- Point C to D consists of a \( 25 \Omega \) resistor.
- Points A to D are linearly arranged in a series connection.

**Diagrams and Explanation:**

The diagram is a series electrical circuit consisting of a single path for current flow:
- Battery of \( 60 \text{ V} \) with \( 2 \Omega \) internal resistance.
- \( 20 \Omega \rightarrow 30 \Omega \rightarrow 25 \Omega \)

**Steps to Calculate Potential Difference:**

To calculate the potential difference across specific points in the circuit:

1. **Find the total resistance in the circuit:**
   \[
   R_{total} = R_{internal} + R_1 + R_2 + R_3 = 2 \Omega + 20 \Omega + 30 \Omega + 25 \Omega = 77 \Omega
   \]

2. **Calculate the total current flowing through the circuit:**
   \[
   I = \frac{V_{battery}}{R_{total}} = \frac{60 \text{ V}}{77 \Omega} \approx 0.779 \text{ A}
   \]

3. **Calculate the potential difference across each resistor using Ohm’s Law (V = IR):**
   - For the \( 20 \
Transcribed Image Text:**Understanding Potential Difference in a Series Circuit** **Objective:** To calculate the potential difference \( V_A - V_B \) and \( V_C - V_D \) in the given circuit. Note that the battery has an internal resistance. **Circuit Description:** The circuit comprises the following components: 1. A battery with an electromotive force (emf) of \( 60 \text{ V} \) and an internal resistance of \( 2 \Omega \). 2. Three resistors connected in series: - \( 20 \Omega \) resistor between points A and B, - \( 30 \Omega \) resistor between points B and C, - \( 25 \Omega \) resistor between points C and D. **Detailed Description:** - The series circuit begins with the positive terminal of the battery. - The negative terminal of the battery is connected through an internal resistance of \( 2 \Omega \). - Point A to B consists of a \( 20 \Omega \) resistor. - Point B to C consists of a \( 30 \Omega \) resistor. - Point C to D consists of a \( 25 \Omega \) resistor. - Points A to D are linearly arranged in a series connection. **Diagrams and Explanation:** The diagram is a series electrical circuit consisting of a single path for current flow: - Battery of \( 60 \text{ V} \) with \( 2 \Omega \) internal resistance. - \( 20 \Omega \rightarrow 30 \Omega \rightarrow 25 \Omega \) **Steps to Calculate Potential Difference:** To calculate the potential difference across specific points in the circuit: 1. **Find the total resistance in the circuit:** \[ R_{total} = R_{internal} + R_1 + R_2 + R_3 = 2 \Omega + 20 \Omega + 30 \Omega + 25 \Omega = 77 \Omega \] 2. **Calculate the total current flowing through the circuit:** \[ I = \frac{V_{battery}}{R_{total}} = \frac{60 \text{ V}}{77 \Omega} \approx 0.779 \text{ A} \] 3. **Calculate the potential difference across each resistor using Ohm’s Law (V = IR):** - For the \( 20 \
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