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?

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Answer 1 · 2015-09-27T22:03:16

$2.4 V$ $2.4"V"$

farside.ph.utexas.edu

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

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

$C = \frac{Q}{V}$ $C=Q/V$

So:

$Q = C V$ $Q=CV$

Let $C_{1} = 330 pF$ $C_1=330"pF"$

and $C_{2} = 220 pF$ $C_2=220"pF"$

So:

$C_{1} V_{1} = C_{2} V_{2}$ $C_1V_1=C_2V_2$ $(1)$ $" "color(red)((1))$

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

$V_{1} + V_{2} = 6$ $V_1+V_2=6$ $(2)$ $" "color(red)((2))$

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

$V_{2} = (6 - V_{1})$ $V_2=(6-V_1)$

We can now substitute this expression for $V_{2}$ $V_2$ into $(1) \Rightarrow$ $color(red)((1))rArr$

$C_{1} V_{1} = C_{2} (6 - V_{1})$ $C_1V_1=C_2(6-V_1)$

$C_{1} V_{1} = 6 C_{2} - C_{2} V_{1}$ $C_1V_1=6C_2-C_2V_1$

$C_{1} V_{1} + C_{2} V_{1} = 6 C_{2}$ $C_1V_1+C_2V_1=6C_2$

$V_{1} (C_{1} + C_{2}) = 6 C_{2}$ $V_1(C_1+C_2)=6C_2$

$V_{1} = \frac{6 C_{2}}{(C_{1} + C_{2})}$ $V_1=(6C_2)/((C_1+C_2))$

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

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

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