Question

5 capacitors, C1 = 1 x 10^-6 F, C2= 2x 10^-6 F, C3 = 3 x 10^-6 F, C4 = 4 x 10^-6 F, C5= 5 x 10^-6 F, are connected in series-parallel combination to a 100 volt source as shown in the figure below. Calculate a. total capacitance (CT) b. total charge (QT) c. the individual charges (Q1, Q2, Q3, Q4, Q5) and the individual voltages (V1, V2, V3, V4, V5).

Answer 1

We will assume that the array for these capacitors connected in a series-parallel combination is:

First, we simply add the capacitors to obtain the equivalent circuit, the sum of the parallel capacitors C₂ and C₃ will give the equivalent capacitor C₂₃ and so on with capacitors C₃ and C₄ to give C₄₅:

$C_{23}=C_2+C_3 =(2+3)\mu F = 5\, \mu F$

$C_{45}=C_4+C_5 =(4+5)\mu F = 9\, \mu F$

At last, we find that the equivalent circuit has a total capacitance C_T equal to 762.7 mF

$C_{T}=[C_{1}+C_{23}+C_{45}]^{-1} =[\frac{1}{1\mu F}+\frac{1}{5\mu F}+\frac{1}{9\mu F}]^{-1}$

$C_{T}=[\frac{59}{45\mu F}]^{-1}=\frac{45}{59}\,\mu F \approx \, 762.71 \,mF$

I'll describe the following steps: since we know the equivalent capacitance we can calculate the total charge that will also be the equivalent charge for the capacitors C₁, C_23, and C₄₅ (due to the series array they have). By using V = Q/C we find the voltages V₁, V_23, and V₄₅. Finally, the capacitors that are connected in a parallel array will have the same voltage V and since we know the capacitances we can calculate the charge on each one of them. All of this is resumed here:

$V_T=\frac{Q_T}{C_T} \to Q_{T}=V_{T}C_{T}=(100\,V)(\frac{45}{59}\mu F)=76.271\,\mu C$

$Q_T=76.271\,\mu C=Q_1=Q_{23}=Q_{45}$

$V_1=\frac{Q_1}{C_1}=\frac{76.271\,\mu C}{1\mu F}=76.271\,V$

$V_{23}=V_2=V_3=\frac{Q_{23}}{C_{23}}=\frac{76.271\,\mu C}{5\mu F}=15.254\,V$

$V_{45}=V_4=V_5=\frac{Q_{45}}{C_{45}}=\frac{76.271\,\mu C}{9\mu F}=8.475\,V$

$Q_2=C_2V_2=(15.254\,V)(2\mu F)=30.508\,\mu C$

$Q_3=C_3V_3=(15.254\,V)(3\mu F)=45.763\,\mu C$

$Q_4=C_4V_4=(8.475\,V)(4\mu F)=33.898\,\mu C$

$Q_5=C_5V_5=(8.475\,V)(5\mu F)=42.373\,\mu C$

In conclusion,

a) C_T = 762.71 mF

b) Q_T = 76.271 $\mu C$

c) Q₁ = 76.271 $\mu C$ ; V₁ = 76.271 V

c) Q₂ = 30.508 $\mu C$ ; V₂ = 15.254 V

c) Q₃ = 45.763 $\mu C$ ; V₃ = 15.254 V

c) Q₄ = 33.898 $\mu C$ ; V₄ = 8.475 V

c) Q₅ = 42.373 $\mu C$ ; V₅ = 8.475 V

Reference:

• Young, H. D., & Freedman, R. A. (2014). Sears and Zemansky's University Physics (with Modern Physics Technology Update). Edinburgh: Pearson.