A constant current of 913 mA flows through a light bulb. During a 45-minute period, the light bulb gives off 61 kJ of energy in the form of light and heat. Calculate the voltage drop across the bulb.

To find the voltage drop across a light bulb with a current of 913 mA and 61 kJ of energy given off in 45 minutes, calculate the power and then use the power formula to find the voltage, resulting in approximately 24.75 V.

Explanation:

To calculate the voltage drop across a light bulb, where a constant current of 913 mA (0.913 A) flows through it and 61 kJ (61,000 J) of energy is given off during a 45-minute period, we use the formula for power (P=IV) and the relationship between power and energy (P=E/t). First, convert the time from minutes to seconds: 45 minutes = 2700 seconds. Next, knowing that P=E/t, we find the power to be P = 61000 J / 2700 s = 22.59 W. Using the power formula (P=IV), rearrange to solve for voltage (V): V=P/I. Plug in the values: V = 22.59 W / 0.913 A, resulting in a voltage drop of approximately 24.75 V across the light bulb.