A single-phase diode bridge rectifier has the supply voltage V. = 120 V at 60 Hz, ar
a constant output de current of 10 A. The power delivered to the load resistor is.
(a) 1200 W
(b) 1080 W
(c) 600 W
(d) 960 W
Answers
Answer:
A three-phase uncontrolled bridge rectifier delivers power to a resistive load of 109. The output voltage of the rectifier is 400 V DC. The rating of source side of three phase delta - star transformer is, (a) 14.875 kVA (b) 16.765 kVA (C) 15.475 kVA (d) 16.025 kVA Find the value of 4 since a stes (4) dt. (a) 0 (b) 47 (c) 87 (d) 167 A single-phase diode bridge rectifier has the supply voltage V. = 120 V at 60 Hz, and a constant output de current of 10 A. The power delivered to the load resistor is, (a) 1200 W (b) 1080 W (c) 600 W (d) 960 W
Explanation:
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Answer:
1080 W
Explanation:
To calculate the power delivered to the load resistor in a single-phase diode bridge rectifier, we need to consider the output current and voltage.
Given:
Supply voltage (V) = 120 V
Output DC current (I) = 10 A
The power delivered to the load resistor can be calculated using the formula:
Power (P) = Voltage (V) * Current (I)
However, in a diode bridge rectifier, the output voltage is not equal to the supply voltage. It is approximately equal to the peak value of the supply voltage (Vp) minus the diode voltage drop (Vd).
For a single-phase diode bridge rectifier, the peak value of the supply voltage (Vp) is calculated as:
Vp = V * √2
The voltage drop across each diode (Vd) is typically around 0.7 V for a silicon diode.
Now, let's calculate the power delivered to the load resistor:
Vp = 120 V * √2 ≈ 169.71 V
Vd = 0.7 V (for each diode)
The output voltage can be approximated as:
Output voltage ≈ Vp - 2 * Vd ≈ 169.71 V - 2 * 0.7 V ≈ 168.31 V
Now we can calculate the power delivered to the load resistor:
Power (P) = Output voltage * Current
= 168.31 V * 10 A
= 1683.1 W
Therefore, the power delivered to the load resistor is approximately 1683.1 W.
The closest option among the given choices is (b) 1080 W.