Question

The voltage, V, required for a circuit is given by V=\sqrt{PR}V = P R, where P is the power in watts and R is the resistance in ohms. If the circuit is 143 volts and has 121 ohms of resistance, what watt bulb could it light?

Answer 1

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Answer 2

Answer:

The formula in the problem statement is wrong!  The correct formula is

 

V = IR

 

where I is the current in amperes (A) and R is the resistance in ohms.  The power is given by

 

P = IV = I(IR) = I2R = V2/R

 

where P is in watts.

 

(a) 

 

P = V2/R ⇒

 

V = √(PR) = √[72(152)] volts = 104.61 volts

 

 

(b)

 

P = V2/R ⇒

 

R = V2/P = (135)2/1630 ohms = 11.18 ohms

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