what is the voltage across C1?

[charmcaster.engg]

If C1, a 4.7

F capacitor, and C2, a 3.3

F capacitor are in series with 18 Vdcapplied, what is the voltage across C1?
Frequency is not mentioned.

 

[davenn]

Frequency is not mentioned.

because DC voltage doesn't have a frequency

google capacitors in series

 

[charmcaster.engg]

because DC voltage doesn't have a frequency

google capacitors in series

Then how to find Xc?

 

[davenn]

There is NO Xc ( capacitive reactance) ..... again ... Its DC voltage

did you google voltage across capacitors in series ?

 

[charmcaster.engg]

There is NO Xc ( capacitive reactance) ..... again ... Its DC voltage

did you google voltage across capacitors in series ?

V=Vc1+Vc2 and Vc1=Q/C1, Vc2=Q/C2. I find out total capacitance using C1*C1/C1+C2=1.94uF. Then Q=C*V=34.92. n then Vc1=Q/C1=7.42.
Am I right?

 

[Laplace]

Correct, but there is no need to calculate the charge Q. Just recognize that the charge is equal in both capacitors. C=Q/V, Q1=Q2, C1∙V1=C2∙V2. The voltage law gives V1+V2=18.

So just solve these two equations for V1: C1∙V1=C2∙V2 & V1+V2=18

V2=(C1/C2)V1, (1+C1/C2)V1=18, V1=18/(1+4.7/3.3) = 7.425

 

[charmcaster.engg]

Correct, but there is no need to calculate the charge Q. Just recognize that the charge is equal in both capacitors. C=Q/V, Q1=Q2, C1∙V1=C2∙V2. The voltage law gives V1+V2=18.

So just solve these two equations for V1: C1∙V1=C2∙V2 & V1+V2=18

V2=(C1/C2)V1, (1+C1/C2)V1=18, V1=18/(1+4.7/3.3) = 7.425

Thanks!