a device is marked as 150w,220v,if the supply voltage is reduced 110v .what will be its power?
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Explanation:
HEY MATE HERE IS YOUR ANSWER
The power dissipated by an electric bulb is given as P = VI.
The power dissipated by an electric bulb is given as P = VI.In this case, Power P = 100 W and Voltage V = 220 V.
The power dissipated by an electric bulb is given as P = VI.In this case, Power P = 100 W and Voltage V = 220 V.So, current I is I = P/V = 100/220 = 0.45 A.
The power dissipated by an electric bulb is given as P = VI.In this case, Power P = 100 W and Voltage V = 220 V.So, current I is I = P/V = 100/220 = 0.45 A.When the supplied voltage is 110 V, the power consumed will be P=VI = 110 * 0.45 = 50 W.
The power dissipated by an electric bulb is given as P = VI.In this case, Power P = 100 W and Voltage V = 220 V.So, current I is I = P/V = 100/220 = 0.45 A.When the supplied voltage is 110 V, the power consumed will be P=VI = 110 * 0.45 = 50 W.Hence, the power consumed when operated at 110 V is 50 W.
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Hey! Here is your answer..<❤>
❇If the voltage of 110 Volts is applied to it on full load it will draw a current of 150/110 Amps current which is more than rated current for which the instrument is designed obviously it will get damaged due to load current
❇The device is designated by the manufacturer to consume 150W at 230V,if you lower the voltage (assuming there are connections you can change on the device to do so) you will increase its current draw. P = VI so I = P/V at 230V I = 150/230 or 652 mA. If you lower the voltage to 110V you will have 150/110 or 1364 mA or 1.364A
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