Two conductors made of the same material are connected across the same potential difference. Conductor A has twice the diameter and twice the length of conductor B. What is the ratio of the power delivered to A to the power delivered to B?
Two conductors made of the same material are connected across the same potential difference. Conductor A has twice the diameter and twice the length of conductor B. What is the ratio of the power delivered to A to the power delivered to B?
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![**Problem Statement:**
Two conductors made of the same material are connected across the same potential difference. Conductor A has twice the diameter and twice the length of conductor B. What is the ratio of the power delivered to A to the power delivered to B?
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**Solution Explanation:**
When two conductors are made from the same material, they have the same resistivity. The resistance \( R \) of a conductor can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
Since Conductor A has twice the diameter of Conductor B, its radius is also twice, and therefore, its cross-sectional area is four times that of B because area \( A \) for a cylinder is \( \pi r^2 \).
Given:
- Length of A is \( 2L_B \)
- Diameter of A is \( 2d_B \) which makes its area \( 4A_B \)
Thus, the resistance of A, \( R_A \), becomes:
\[ R_A = \frac{\rho \cdot 2L_B}{4A_B} = \frac{\rho \cdot L_B}{2A_B} = \frac{1}{2}R_B \]
The power \( P \) delivered to a conductor connected across a potential difference \( V \) is given by:
\[ P = \frac{V^2}{R} \]
Therefore, the power delivered to each conductor will be:
\[ P_A = \frac{V^2}{R_A} \]
\[ P_B = \frac{V^2}{R_B} \]
Since \( R_A = \frac{1}{2}R_B \),
\[ P_A = \frac{V^2}{\frac{1}{2}R_B} = 2 \times \frac{V^2}{R_B} = 2P_B \]
**Conclusion:**
The ratio of the power delivered to A to the power delivered to B is 2:1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c4a5f83-7aca-4e4e-8516-6de83483f29d%2Fcd34667f-7250-484d-b2e1-5db0015a3c25%2Fhojv12j_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Two conductors made of the same material are connected across the same potential difference. Conductor A has twice the diameter and twice the length of conductor B. What is the ratio of the power delivered to A to the power delivered to B?
---
**Solution Explanation:**
When two conductors are made from the same material, they have the same resistivity. The resistance \( R \) of a conductor can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
Since Conductor A has twice the diameter of Conductor B, its radius is also twice, and therefore, its cross-sectional area is four times that of B because area \( A \) for a cylinder is \( \pi r^2 \).
Given:
- Length of A is \( 2L_B \)
- Diameter of A is \( 2d_B \) which makes its area \( 4A_B \)
Thus, the resistance of A, \( R_A \), becomes:
\[ R_A = \frac{\rho \cdot 2L_B}{4A_B} = \frac{\rho \cdot L_B}{2A_B} = \frac{1}{2}R_B \]
The power \( P \) delivered to a conductor connected across a potential difference \( V \) is given by:
\[ P = \frac{V^2}{R} \]
Therefore, the power delivered to each conductor will be:
\[ P_A = \frac{V^2}{R_A} \]
\[ P_B = \frac{V^2}{R_B} \]
Since \( R_A = \frac{1}{2}R_B \),
\[ P_A = \frac{V^2}{\frac{1}{2}R_B} = 2 \times \frac{V^2}{R_B} = 2P_B \]
**Conclusion:**
The ratio of the power delivered to A to the power delivered to B is 2:1.
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