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A bulb connected to 50 V d.c. consumes 20 W power. Then the bulb is connected to a capacitor in an a.c. power supply of 250 V, 50 Hz. Find the value of the capacitor required so that the bulb drows the same amount of current. Data : P = 20 W, V = 50 V, v = 50 Hz, C = ? |
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Answer» <P> Solution :`P = VI``:. I = (P)/(V) = (20)/(50) = 0.4A` `:.` Resistance, `R = (V)/(I) = (50)/(0.4) = 125 Omega`  The IMPEDANCE, `Z = (V)/(I) = (250)/(0.4) = 625 Omega` `:. Z = SQRT(R^(2) + ((1)/(omega C))^(2)) = sqrt(R^(2) + ((1)/(2pi v C))^(2))` |
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