Question

A metal -filament lamp rated at 750W 100V is to be connected in series with a capacitance across a 230V 50Hz supply.Calculate the value of capacitance required. Draw phasor diagram.​

Answer 1

Answer:

pls ask the brainliest person if you want the answer

Answer 2

Answer:

The correct value of the capacitance required will be 96μF.

Explanation:

Let, R should stand-in for the lamp. While the voltage VC across C lags I by 90°, the voltage VR across R is in phase with the current I. The phasor sum of VR and VC yields the final voltage, V.

V² = VR² + VC²

(230)² = (100)² + VC²

VC = 270V

Given that rated current of the lamp = $\frac{750}{100}$

= 7.5A

From equation = 2πfLIm sin(2πft + $\frac{\pi }{2}$)

7.5 = 2 × 3.14 × 60 × C × 207

C = 96 × 10⁻⁶F

C = 96μF

The ratio of the maximum charge Q that may be stored in a capacitor to the applied voltage V across its plates is known as the capacitance C of a capacitor. In other terms, according to the formula C = QV, capacitance is the maximum charge that can be held on a device per volt.

Thus, a capacitor's capacitance measures how much electricity it can store per unit of the voltage applied across its plates.