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The Voltage drop across a 500 Watt electric heater having resistance of 5 Ohm is?
ZeeshanMuhammad
Two equations for power are: P=I2R and P=EI where P=power, E=voltage, and I=current.
Solving the first equation for I and the second equation for E gives: I=(P/R)1/2 E=P/I
Now substitute the first equation into the second.
E=P/(P/R)1/2 = (P1/2)(R1/2)
Using the given values:
E = (5001/2)(51/2) = ((5)(100))1/2(51/2) = (51/2)(10)(51/2) = 50 volts
ANSWER: WHATEVER THE SOURCE VOLTAGE MINUS THE IR DROP OF WIRES
Wiki User
∙ 13y ago
Abdullah Sadiqui
50 volt
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A cell of emf 3.0 volts and internal resistance of 0.01 ohm is connected through an ammeter of 0.05-ohm internal resistance to a 5.0 rheostat by wire having a total resistance of 0.85-ohm?
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A cell of emf 3.0 volts and internal resistance of 0.01 ohm is connected through an ammeter of 0.05-ohm internal resistance to a 5.0 rheostat by wire having a total resistance of 0.85-ohm?
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Number of bulbs . . . Load voltage . . . Load current 0 . . . . . . . . . . 120 V . . . . . . . . 0 A. 1 . . . . . . . . . . 117.65 V .. . . . . 23.53 A. 2 . . . . . . . . . . 115.38 . . . . . . . 46.15 3 . . . . . . . . . . 113.21 . . . . . . . 67.92 4 . . . . . . . . . . 111.11 . . . . . . . 88.88 5 . . . . . . . . . . 109.09 . . . . . . . 109.09
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