Question

An ac generator has a maximum emf output of 155 v.

a. find the rms emf output.
b. find the rms current in the circuit when the generator is connected to a 53 ω resistor.

Answer 1

a) For an ac generator, the rms value of the voltage is given by:

`V_(rms) = (V_0)/( √(2) )`
where
`V_0` is the peak value of the voltage.

For the generator in our problem, the peak value is
`V_0=155 V`, therefore the rms value of the voltage is

`V_(rms)= (V_0)/( √(2) )= (155 V)/( √(2) )=109.6 V`


b) We can find the rms current in the circuit by using Ohm's law, which is also valid when using the rms values of the current and the voltage:

`I_(rms) = (V_(rms))/(R)`
where R is the resistance conncected to generator. If we use
`R=53 \Omega`, we find

`I_(rms) = (V_(rms))/( R )= (109.6 V)/(53 \Omega) =2.07 A`

Answer 2

Vp = 155 V
R = 53 ω

(a) RMS emf output

Vrms = 0.7071Vp = 0.7071*155 = 109.6 V

(b) RMS current
From ohm's law,
Vrms = IrmsR => Irms = Vrms/R = 109.6/53 = 2.07 Amps