Capacitors Concept Page - 4

Definition

Infinite capacitors in series and parallel

The equivalent capacity of the infinite network shown in the figure (across AB) is (Capacity of each capacitor is   1 μ$\mu$ F) to be found out.
If we remove one set from infinite it is still equal to infinite series, assume the equivalent capacitance is Ceff$C_{eff}$, 
now 1μF$1\mu F$ and Ceff$C_{eff}$ are in parallel, their equivalent capacitance is Ceff+1$C_{eff}+1$.
Now 2 capacitors of 1μF$1\mu F$ and Ceff+1$C_{eff}+1$ are in series.
1Ceff=11+11+11+Ceff${\displaystyle \frac{1}{C_{eff}}}={\displaystyle \frac{1}{1}}+{\displaystyle \frac{1}{1}}+{\displaystyle \frac{1}{1+C_{eff}}}$
1Ceff=2+11+Ceff${\displaystyle \frac{1}{C_{eff}}}=2+{\displaystyle \frac{1}{1+C_{eff}}}$
1Ceff=2+2Ceff+11+Ceff${\displaystyle \frac{1}{C_{eff}}}={\displaystyle \frac{2+2C_{eff}+1}{1+C_{eff}}}$
1+Ceff=3Ceff+2C2eff$1+C_{eff}=3C_{eff}+2C_{eff}^2$
2C2eff+2Ceff−1=0$2C_{eff}^2+2C_{eff}-1=0$
⇒Ceff=3√−12$\Rightarrow C_{eff}={\displaystyle \frac{\sqrt{3}-1}{2}}$

Example

Capacitor circuit with symmetry

Example: For the given capacitor circuit in figure (a), find the equivalent capacitance across A and B.
Solution:
The given capacitor is symmetric about the line AB. 
∴VD=VF,VE=VG$\therefore V_D=V_F,V_E=V_G$
Also, H is the middle point of the circuit and remains so if the node H is split into H1$H_1$ and H2$H_2$.
Hence, the circuit can be redrawn as shown in figure (b). It is now a simple series parallel circuit.
Capacitance along the path AH1B$A{H}_{1}B$ is C1=C2$C_1={\displaystyle \frac{C}{2}}$

Capacitance along the path FH2G$F{H}_{2}G$ is C2=2C2=C$C_2={\displaystyle \frac{2C}{2}}=C$ 

Capacitance between FG$FG$ is C3=2C+C=3C$C_3=2C+C=3C$

Capacitance along the path AFGB$AFGB$ is C4$C4$

1C4=12C+13C+12C${\displaystyle \frac{1}{C_4}}={\displaystyle \frac{1}{2C}}+{\displaystyle \frac{1}{3C}}+{\displaystyle \frac{1}{2C}}$
C4=3C4$C_4={\displaystyle \frac{3C}{4}}$
Capacitance between AB is Ceq=C2+3C4$C_{eq}={\displaystyle \frac{C}{2}}+{\displaystyle \frac{3C}{4}}$
C=5C4$C={\displaystyle \frac{5C}{4}}$

Note:
Due to symmetry,
VAH=VHB$V_{AH}=V_{HB}$
VAD=VAF=VEB=VGB$V_{AD}=V_{AF}=V_{EB}=V_{GB}$ 
VDH=VHE=VFH=VFG$V_{DH}=V_{HE}=V_{FH}=V_{FG}$
VDE=VFG$V_{DE}=V_{FG}$
Assigning charge to all capacitors in accordance to above equations and summing up the voltages along each path, charge on all capacitors can be found. This is an alternative method to find equivalent capacitance of the circuit.

Example

Wheatstone bridge in capacitors

Example:
Find the effective capacitance between the point P and Q in the given figure .
Solution:The given circuit can be rebuilt as shown in the figure (b). This is a wheatstone bridge for capacitors. The capacitor in the middle can thus be neglected.
Cnet=4μF$C_{net}=4\mu F$

Definition

Star and delta configurations

c1=C2C3+C3C1+C1C2C1$c_1={\displaystyle \frac{C_2{C}_3+C_3{C}_1+C_1{C}_2}{C_1}}$
c2=C2C3+C3C1+C1C2C2$c_2={\displaystyle \frac{C_2{C}_3+C_3{C}_1+C_1{C}_2}{C_2}}$
c3=C2C3+C3C1+C1C2C3$c_3={\displaystyle \frac{C_2{C}_3+C_3{C}_1+C_1{C}_2}{C_3}}$
C1=c2c3c1+c2+c3$C_1={\displaystyle \frac{c_2{c}_3}{c_1+c_2+c_3}}$
C2=c3c1c1+c2+c3$C_2={\displaystyle \frac{c_3{c}_1}{c_1+c_2+c_3}}$
C3=c1c2c1+c2+c3$C_3={\displaystyle \frac{c_1{c}_2}{c_1+c_2+c_3}}$

Example

Energy lost on connecting two capacitors

Example:
A condenser of capacity 10μF$10\mu F$ is charged to a potential of 500 V$500V$. Its terminals are then connected to those of an uncharged condenser of capacity 40μF$40\mu F$. Find the loss of energy in connecting them.Solution:
Initial Energy  =12CV2=1210×10−6×500×500=1.25J$={\displaystyle \frac{1}{2}}C{V}^2={\displaystyle \frac{1}{2}}10\times {10}^{-6}\times 500\times 500=1.25J$
Total Charge =CV=10×10−6×500=5×10−3$=CV=10\times {10}^{-6}\times 500=5\times {10}^{-3}$
After connecting the 2 capacitors , Charge remains same . 
Effective capacitance =C1+C2=50×10−6$=C_1+C_2=50\times {10}^{-6}$
Voltage across the 2 must be equal . 
5×10−3=50×10−6×V$5\times {10}^{-3}=50\times {10}^{-6}\times V$
V=100V$V=100V$
Final Energy Stored =1250×10−6×100×100=0.25J$={\displaystyle \frac{1}{2}}50\times {10}^{-6}\times 100\times 100=0.25J$
Hence Energy Lost = Initial Energy stored - Final Energy Stored =1.25J−0.25J=1J$=1.25J-0.25J=1J$

Definition

Polarisation

When an electric field is applied there is small displacement of bound charges creating small  electric dipoles within the dielectric. This phenomenon is called Polarization.
The polarisation of a region is defined as the dipole moment per unit volume.
For linear isotropic dielectrics, it is given by 
P=χeE$P={\chi }_{e}E$
where χe${\chi }_e$ is the dielectric susceptibility of the dielectric medium

Definition

Electric susceptibility

Polarisation in a dielectric is given by: P=χeE$P={\chi }_{e}E$
where
χ:$\chi :$ Electric susceptibility of the medium
E:$E:$ External electric field
Electric susceptibility is related to the molecular properties of the substance.
It is also given by:
χ=εo(K−1)$\chi ={\epsilon }_o(K-1)$
where K:$K:$ dielectric constant of the medium

Definition

Effect of electric field on dielectric material

Dielectric are non-conducting substances and have negligible number of charge carriers.
When a conductor is placed in an external electric field, free charges move such that the net field inside the conductor is zero. This is not possible in a dielectric.
In a dielectric, the electric field induces dipole moment by reorienting the molecules of the dielectric. This causes an electric field opposing the external electric field. Hence, the electric field inside a dielectric is smaller than the original electric field. 

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