Chapter10: Direct-current Circuits
Section: Chapter Questions
Problem 72AP: The rather simple circuit shown below is known as a voltage divider. The symbol consisting of three...
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Please answer the question listed below and select the correct answer choice below. Thank you!
![**Question 21:** What is the voltage drop across the 20-ohm resistor in the circuit?
**Options:**
- A. 6 V
- B. 4 V
- C. 2 V
- D. 0 V
**Circuit Diagram Explanation:**
The diagram shows a circuit with a 6V battery connected across a parallel combination of two resistors, each with a resistance of 10 ohms. This parallel section is then connected in series with a 20-ohm resistor.
To solve for the voltage drop across the 20-ohm resistor, we need to consider the total equivalent resistance of the circuit and the voltage distribution according to Ohm's Law.
1. **Find equivalent resistance of the parallel section**:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \quad \Rightarrow \quad R_{\text{parallel}} = 5 \, \Omega
\]
2. **Find total resistance in the circuit**:
\[
R_{\text{total}} = R_{\text{parallel}} + R_{20 \, \Omega} = 5 \, \Omega + 20 \, \Omega = 25 \, \Omega
\]
3. **Calculate current using Ohm's Law**:
\[
I = \frac{V}{R_{\text{total}}} = \frac{6 \, \text{V}}{25 \, \Omega} = 0.24 \, \text{A}
\]
4. **Calculate voltage drop across the 20-ohm resistor**:
\[
V_{\text{20} \, \Omega} = I \times R_{20 \, \Omega} = 0.24 \, \text{A} \times 20 \, \Omega = 4.8 \, \text{V}
\]
Therefore, the closest correct answer is:
- **B. 4 V**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F281e2029-de70-4b7c-8ebc-bf0179904e66%2Fe7ec8376-ad91-4422-8c14-38c9fc8a484f%2Fhizstjo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 21:** What is the voltage drop across the 20-ohm resistor in the circuit?
**Options:**
- A. 6 V
- B. 4 V
- C. 2 V
- D. 0 V
**Circuit Diagram Explanation:**
The diagram shows a circuit with a 6V battery connected across a parallel combination of two resistors, each with a resistance of 10 ohms. This parallel section is then connected in series with a 20-ohm resistor.
To solve for the voltage drop across the 20-ohm resistor, we need to consider the total equivalent resistance of the circuit and the voltage distribution according to Ohm's Law.
1. **Find equivalent resistance of the parallel section**:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \quad \Rightarrow \quad R_{\text{parallel}} = 5 \, \Omega
\]
2. **Find total resistance in the circuit**:
\[
R_{\text{total}} = R_{\text{parallel}} + R_{20 \, \Omega} = 5 \, \Omega + 20 \, \Omega = 25 \, \Omega
\]
3. **Calculate current using Ohm's Law**:
\[
I = \frac{V}{R_{\text{total}}} = \frac{6 \, \text{V}}{25 \, \Omega} = 0.24 \, \text{A}
\]
4. **Calculate voltage drop across the 20-ohm resistor**:
\[
V_{\text{20} \, \Omega} = I \times R_{20 \, \Omega} = 0.24 \, \text{A} \times 20 \, \Omega = 4.8 \, \text{V}
\]
Therefore, the closest correct answer is:
- **B. 4 V**
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