Question

A 4.0 x 10^(-6) F capacitor is connected in series with a 3.0 x 10^(-6) F capacitor. This series combination is connected in parallek with a 6.0 x 10^(-6) F capacitor. The whole arrangement is then connected in series with a 12.0 V battery.
(i) Draw a curcuit diagram of the arrangement described above
(ii) Calculate the total charge in stored in the circuit and hence calculate the charge stored in each capacitor.

Answer 1

(a) The curcuit diagram of the arrangement described above:

(b) The total charge stored in circuit:

$q=CV_b,$

where C - the equivalent capacitance of the circuit:

$C=\frac{C_1C_2}{C_1+C_2}+C_3.$

So, $q=V_b(\frac{C_1C_2}{C_1+C_2}+C_3)=12(\frac{4\cdot 10^{-6}\cdot 3\cdot 10^{-6}}{4\cdot 10^{-6}+3\cdot 10^{-6}}+6\cdot 10^{-6})\approx 9.3\cdot 10^{-5}\space C.$

The charge stored in C3:

$q_3=C_3V_b=6\cdot 10^{-6}\cdot12=7.2\cdot 10^{-5}\space C.$

The charges stored in two capacitors C1 and C2, connected in series:

$q_1=q_2=\frac{q_{12}}{2},$

$q_{12}=C_{12}V_b,$

where $C_{12}$ - the equivalent capacitance of two capacitors, connected in series:

$C_{12}=\frac{C_1C_2}{C_1+C_2}.$

So, $q_1=q_2=\frac{V_bC_1C_2}{2(C_1+C_2)}=\frac{12\cdot 4\cdot 10^{-6}\cdot 3\cdot 10^{-6}}{2(4\cdot 10^{-6}+3\cdot 10^{-6})}\approx 1.05\cdot 10^{-5}\space C.$

Answer: the total charge is $9.3\cdot 10^{-5}\space C$; the charges stored in each capacitor: $q_1=q_2=1.05\cdot 10^{-5}\space C$, $q_3=7.2\cdot 10^{-5}\space C$.