Question
Download Solution PDFSuppose in a circuit, a silver wire of length L and cross-sectional area A is replaced by an aluminium wire of length 5 L and cross-sectional area 9 A. The resistance of the circuit will ________.
(Given that ρsilver = 1.6 × 10-8 Ω m and ρAluminium = 2.6 × 10-8 Ωm)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe correct answer is decrease to 0.9 times of itself.
Key Points
- Resistance
- There is a relationship between the potential difference across a conductor and the current that flows through it, according to Ohm's law. It comes from,
V ∝ I V = IR
Where V is the potential difference measured across the conductor in volts, I is the conductor's current in amperes, and R is the resistance constant in ohms. - The following factors influence a conductor's electrical resistance:
- The cross-sectional area of the conductor
- Length of the conductor
- The material of the conductor
- The temperature of the conducting material
- Electrical resistance is inversely proportional to the conductor's cross-sectional area (A) and directly proportional to its length (L). The relationship that follows provides it.
R= ρ L/A
- There is a relationship between the potential difference across a conductor and the current that flows through it, according to Ohm's law. It comes from,
Additional Information
- Calculation
- Given,
A silver wire of length L and cross-sectional area A is replaced by an aluminum wire of length 5 L and cross-sectional area 9 A.
ρsilver = 1.6 × 10-8 Ω m and ρAluminium = 2.6 × 10-8 Ωm
We know,
R= ρ L/A
Rsilver= 1.6 × 10-8 L/A
Raluminium= 2.6 × 10-8 × 5L/9A
Rsilver/Raluminum= 1.6× 9/ 2.6× 5= 1.107
Ralluminium/Rsilver= 0.90
- Given,
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