A resistor is connected first in parallel & then in series with 2.00 ohm resistor?

a battery delivers five times as much current to the parallel combination as it does to the series cobination to the series combination. determine two values for R





2. a light is connected to 120.0 v wall socket. the current in the bulb depends on the time t according to the relation t=(0.707A)sin{(314Hz)t}.


a)what is the frequncy of the alternating current? b) determine the resistance of the bulb's filament.


c) what is the average power consumed by the light bulb

A resistor is connected first in parallel %26amp; then in series with 2.00 ohm resistor?
a) 60 cycles per second (In some countries 50 cycles)


b) Bulb have two resistance values of filament, when are connected is greatest that when is disconnected


c) The formula for the consume power is P= U x I , P is the power, U is the line voltage and I is the consumer amperes of the bulb
Reply:V (r1+2)/r1*2 = 5 (v/(r1+2)


(r1+2)/2r1 = 5/(r1+2)


(r1+2)^2 = 10r1


Solve for r1





I = Imax*sin wt where w = 2*pi*f = 314 (314=100 pi)


so, 2pi*f = 100 pi, so f = 50 Hz.





When you said 120 Volts, I assume it is rms 120 volts.


The rms value of current is I max / sqrt(2)


= I max*0.707 (as 1/sqrt(2) = 0.707


I max = 0.707 so I rms = 1/2 (that is 1/(sqrt2*sqrt2) = 1/2)





v=120, I =1/2 so R= 120*2 =240 ohms.





Power = V*I -- = 60 WATTS (RMS)





Average power = 1.1 times rms.


=66 watts average.