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------------------------------------------------ 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. Determine the following:

a) Power
b) Voltage
c) Resistance
d) Current

To find power (P), we use the formula P = IV, where I is the current and V is the voltage. Given I = 913 mA and time t = 45 minutes = 2700 seconds, we first convert the time to seconds. Then, using the formula P = W/t (where W is the energy in joules), we calculate power. In this case, W = 61 kJ = 61,000 J. So, power (P) = 61,000 J / 2700 s = 45.6 W.

To find voltage (V), we use the formula P = IV. We have P and I, so we can rearrange the formula to solve for V: V = P/I. Plugging in the values, V = 45,600 mW / 913 mA = 41.7 V.

Resistance (R) can be found using Ohm's law: R = V/I. Given V and I, we calculate R: R = 41,700 mV / 913 mA = 45.6 Ω.

Since we're given the constant current (I) of 913 mA, the answer to the question "d) Current" is straightforward: Current = 913 mA. This completes the analysis of the given electrical parameters for the light bulb.