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A single-pulse transformer with secondary voltage of 230 V, 50 Hz, delivers power to bulb of R = 10 Ω through a half-wave controlled rectifier circuit. For α = 60°, find the average current in the bulb
- a)2.5 A
- b)7.7 A
- c)9.6 A
- d)3 A
Correct answer is option 'B'. Can you explain this answer?
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A single-pulse transformer with secondary voltage of 230 V, 50 Hz, del...
First find the average voltage, than Io = Vo/R
Vo = (Vm/2π) x (1+cosα).
Vo = (Vm/2π) x (1+cosα).
Most Upvoted Answer
A single-pulse transformer with secondary voltage of 230 V, 50 Hz, del...
Simplicity, assume that the transformer has no leakage inductance, and the rectifier circuit has no voltage drop. The waveform of the output voltage of the transformer is shown below:
The bulb of R = 10 is connected in series with the load resistor R_load, and the voltage across the load is given by:
V_load = V_sec * D
where V_sec is the secondary voltage of the transformer, and D is the duty cycle of the rectifier circuit. The duty cycle is defined as the ratio of the time when the rectifier is conducting to the period of the input voltage. In this case, the period of the input voltage is 1/50 seconds, or 20 milliseconds.
The output voltage of the rectifier circuit is shown below:
The duty cycle D is the ratio of the time when the voltage is positive to the period of the input voltage. From the waveform, we can see that the voltage is positive for half of the period, or 10 milliseconds. Therefore:
D = 10 ms / 20 ms = 0.5
Substituting this into the equation for V_load, we get:
V_load = 230 V * 0.5 = 115 V
The power delivered to the load is given by:
P_load = V_load^2 / R_load
Substituting R_load = 10 ohms and V_load = 115 V, we get:
P_load = 115^2 / 10 = 1322.5 W
Therefore, the power delivered to the load is 1322.5 watts.
The bulb of R = 10 is connected in series with the load resistor R_load, and the voltage across the load is given by:
V_load = V_sec * D
where V_sec is the secondary voltage of the transformer, and D is the duty cycle of the rectifier circuit. The duty cycle is defined as the ratio of the time when the rectifier is conducting to the period of the input voltage. In this case, the period of the input voltage is 1/50 seconds, or 20 milliseconds.
The output voltage of the rectifier circuit is shown below:
The duty cycle D is the ratio of the time when the voltage is positive to the period of the input voltage. From the waveform, we can see that the voltage is positive for half of the period, or 10 milliseconds. Therefore:
D = 10 ms / 20 ms = 0.5
Substituting this into the equation for V_load, we get:
V_load = 230 V * 0.5 = 115 V
The power delivered to the load is given by:
P_load = V_load^2 / R_load
Substituting R_load = 10 ohms and V_load = 115 V, we get:
P_load = 115^2 / 10 = 1322.5 W
Therefore, the power delivered to the load is 1322.5 watts.
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