A 330 pF capacitor and a 220 pF capacitor in series are each connected across a 6 V dc source. What is the voltage across the 330 pF capacitor?

1 Answer
Sep 27, 2015

Answer:

#2.4"V"#

Explanation:

farside.ph.utexas.edu

For capacitors in series like this the total charge stored by each capacitor is the same.

The capacitance #C#, charge stored #Q# and the voltage #V# are related by:

#C=Q/V#

So:

#Q=CV#

Let #C_1=330"pF"#

and #C_2=220"pF"#

So:

#C_1V_1=C_2V_2# # " "color(red)((1))#

The voltage drop across the 2 capacitors = 6V so we can write:

#V_1+V_2=6# #" "color(red)((2))#

We have 2 simultaneous equations here so from #color(red)((2))# we can write:

#V_2=(6-V_1)#

We can now substitute this expression for #V_2# into #color(red)((1))rArr#

#C_1V_1=C_2(6-V_1)#

#C_1V_1=6C_2-C_2V_1#

#C_1V_1+C_2V_1=6C_2#

#V_1(C_1+C_2)=6C_2#

#V_1=(6C_2)/((C_1+C_2))#

#V_1=(6xx220xxcancel(10^(-12)))/((330+220)xxcancel(10^(-12)))#

#V_1=1320/550=2.4"V"#

  • which is the voltage drop across the #330"pF"# capacitor.

This also means that #V_2=6-2.4=3.6"V"#