Question

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.

Answer 1

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