Calculate Current in a 50 Ohm Resistor from Half Wave Rectified Voltage

In summary, the conversation discusses how to find the average and rms values of the current in a half wave rectified voltage with a given equation, when applied to a 50 ohm resistor. The solution involves using the integral of the sine wave and applying Ohm's law. A helpful hint is also given to calculate the average for a half-cycle of the sine first.
  • #1
cleas001
2
0
im stuck on this question and was wondering if anyone could help point me in the right direction:
a half wave rectified voltage is described by: v(t)=100.sinX
(that X is mean to be theta)
when applied to a 50 ohm resistor find the average and rms values of the current.


thanks an help is much appreciated...
a stuggling student
 
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  • #2
The key word there is average. The average value of any curve equals the area of that curve divided by its length. In order words, take the integral of the sine wave from 0 to pi then divide by the length of the curve which is pi in this case. Once you find the average voltage it is possible to find the average current using ohms law.
 
Last edited:
  • #3
okay i will certainly give that a good try thanks
 
  • #4
cleas001 said:
im stuck on this question and was wondering if anyone could help point me in the right direction:
a half wave rectified voltage is described by: v(t)=100.sinX
No, after an ideal half-wave rectification, you will have v(t) = 100 |sin(x)|. Be sure to average the absolute value version of the excitation. Hint -- You can calculate the average for a half-cycle of the sine first...
 

Related to Calculate Current in a 50 Ohm Resistor from Half Wave Rectified Voltage

1. How do you calculate the current in a 50 Ohm resistor from half wave rectified voltage?

To calculate the current in a 50 Ohm resistor from half wave rectified voltage, you can use Ohm's Law which states that current (I) is equal to voltage (V) divided by resistance (R). In this case, the voltage will be the peak voltage of the half wave rectified waveform and the resistance will be 50 Ohms.

2. What is half wave rectified voltage?

Half wave rectified voltage is a type of electrical signal that has been converted from an alternating current (AC) to a direct current (DC) by only allowing the positive half of the AC waveform to pass through. This results in a waveform with a series of half cycles instead of full cycles.

3. Why is the resistance value important for calculating current in this scenario?

The resistance value is important because it determines the amount of opposition to the flow of current in the circuit. In this scenario, the 50 Ohm resistor is providing a specific amount of resistance to the current flow, which affects the amount of current that will pass through it. Without knowing the resistance value, it would be impossible to accurately calculate the current.

4. Is there a difference in calculating current for a half wave rectified voltage compared to a full wave rectified voltage?

Yes, there is a difference in calculating current for a half wave rectified voltage compared to a full wave rectified voltage. This is because with half wave rectification, only half of the AC waveform is being used, resulting in a lower average voltage and therefore a lower current. With full wave rectification, both the positive and negative halves of the AC waveform are used, resulting in a higher average voltage and current.

5. Can this calculation be used for other types of resistors besides 50 Ohms?

Yes, this calculation can be used for other types of resistors besides 50 Ohms. The only value that would change in the calculation would be the resistance value (R). As long as the resistance value is known, Ohm's Law can be used to calculate the current in any type of circuit or resistor.

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