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alexandr1967 [171]
3 years ago
8

A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.

Physics
1 answer:
kirill [66]3 years ago
4 0

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current I = 1 A

Voltage V = 12 V

For finding the resistance,

  V = IR

  R = \frac{V}{I}

  R = \frac{12}{1}

  R = 12Ω

(B)

For finding power delivered,

  P = I^{2} R

  P = (1) ^{2} \times 12

  P = 12 Watt

(C)

For finding the potential difference,

   V = IR

   V = 5 \times 10^{-3} \times 2

   V = 10 \times 10^{-3}

   V = 0.01 V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

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