Question

When two capacitors are connected in parallel across a 12.4 V rms, 1.12 kHz oscillator, the oscillator supplies a total rms current of 546 mA . When the same two capacitors are connected to the oscillator in series, the oscillator supplies an rms current of 125 mA . What are the values of the two capacitors?

Answer 1

step;1

Given that

voltage V=12.4 v

frequency f=1.12*10^3 Hz

step;2

now we find Ceq connected in parallel

V=I*Xc

12.4=546*10^-3*1/2*3.14*1.12*10^3*Ceq

12.4=77.63*1/Cea

Ceq=6.3

C1+C2=6.3....................(1)

step;3

now we find the Ceq in series connection

12.4=125*10^-3*1/2*3.14*1.12*10^3*Ceq

Ceq=1.44

C1C2/C1+C2=1.44

C1C2=9.1..............................(2)

=>(C1-C2)^2=(c1+c2)^2-4C1C2

=(6.3)^2-4*9.1

=>(C1-C2)^2=3.29

C1-C2=1.814.............................(3)

now we solve eq1and eq3 we get

C1+C2=6.3

C1-C2=1.814

...............................................

2C1=8.114

C1=4.1F

C1 is substituting in equation 1 we get

4.1+C2=6.3

C2=2.2 F

therefore the capacitor C1=4.1 F

C2=2.2 F