A 150 W light bulb is connected to this power source and it glows at its maximum brightness.
Use the information from the graph to calculate the:
.1 period of the wave
.2 frequency of the wave
3 Resistance of the bulb
.4 RMS current through the bulb
.5 The cost of the electricity it uses in 32 hours if 1 kWh costs R2,50
1. To calculate the period of the wave, we need to find the distance between two consecutive peaks (or two consecutive troughs) on the graph. Looking at the graph, we see that the distance between two consecutive peaks is 0.02 s. Therefore, the period of the wave is 0.02 s.
2. The frequency of the wave is the number of waves that pass a certain point in one second. The formula to calculate frequency is: frequency = 1 / period. Using the period calculated above, the frequency of the wave is 1 / 0.02 = 50 Hz.
3. The resistance of the bulb can be calculated using Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). From the graph, we can see that the voltage is 220 V when the light bulb is glowing at its maximum brightness. We also know that the power of the bulb is 150 W. Using the formula for power (P = V * I), we can rearrange this to find the current, which would be 150 W / 220 V = 0.682 A. Therefore, the resistance of the bulb is 220 V / 0.682 A = 322.43 ohms.
4. The RMS current can be calculated using the formula: Irms = Ipeak / √2. From the graph, we can see that the peak current is 1 A. Therefore, the RMS current through the bulb is 1 A / √2 ≈ 0.707 A.
5. The power consumption of the light bulb can be calculated using the formula: Power = (Watts) x (hours) = (150 W) x (32 hours) = 4800 Wh = 4.8 kWh. To calculate the cost of the electricity used, we multiply the power consumption by the cost of 1 kWh: 4.8 kWh x R2.50 = R12.00. Therefore, the cost of the electricity used by the light bulb in 32 hours would be R12.00.