How much power is dissipated by a light bulb if its applied voltage is 12 V and its resistance is 12 Ω?

1. Break Down the Problem

To find the power dissipated by the light bulb, we can use Ohm's Law and the power formula. The relevant parameters are:

• Voltage ($V = 12 \, \text{V}$)
• Resistance ($R = 12 \, \Omega$)

2. Relevant Concepts

The power ($P$) dissipated by a resistor can be calculated using the formula: $P = \frac{V^2}{R}$ where:

• $V$ is the voltage across the resistor,
• $R$ is the resistance.

3. Analysis and Detail

Substituting the known values into the power formula:

$P = \frac{(12 \, \text{V})^2}{12 \, \Omega}$

Calculating the numerator: $(12 \, \text{V})^2 = 144 \, \text{V}^2$

Now substitute back to find $P$: $P = \frac{144 \, \text{V}^2}{12 \, \Omega} = 12 \, \text{W}$

4. Verify and Summarize

The calculations have been verified, confirming that the power dissipated by the light bulb is 12 Watts.

Final Answer

The power dissipated by the light bulb is $12 \, \text{W}$.