Current Electricity Concept Page - 16

Example

Charge on a capacitor in a charging/discharging circuit

Initially capacitor was uncharged and switch was open. Switch is closed at t=0. Ammeter and voltmeter are ideal. [All units are in SI]After long time, current through the circuit will be zero since capacitor will behave as open circuit. Therefore voltmeter will read 10V.
Just after closing the switch, voltage across capacitor will be 0V. Hence current will be i=20+104+2=5A$i={\displaystyle \frac{20+10}{4+2}}=5A$.
Reading of voltmeter just after closing the switch will be 10−2i=0V$10-2i=0V$.
After long time, entire voltage will appear across the capacitor, hence, Q=CV=2X30=60C

Definition

Non-ideal capacitor

If the dielectric material between the plates of a capacitor has finite resistivity as opposed to infinite resistivity in case of an ideal capacitor then there will be a small amount of current flowing between the plates of the capacitor.Lead resistance and plate effects also exist in non-ideal capacitor.

Example

Energy stored in a capacitor in RC circuit

Example:
Consider the circuit shown in the attached figure. Find the energy stored in the capacitor after time t. The initial charge on the capacitor is 0.
Solution:
Given, q(0)=0$q(0)=0$
At steady state, capacitor will be fully charged and hence will have a potential of Vs$V_s$ across it.
Hence, charge stored in C in steady state is q(∞)=CVs$q(\mathrm{\infty })={\displaystyle \frac{C}{V_s}}$
Charge stored in the capacitor after time t is given by:
q(t)=q(∞)−[q(∞)−q(0)]e−tRC${\displaystyle q(t)=q(\mathrm{\infty })-[q(\mathrm{\infty })-q(0)]e^{-\frac{t}{RC}}}$ 
q(t)=CVs(1−e−tRC)${\displaystyle q(t)={\displaystyle \frac{C}{V_s}}(1-e^{-\frac{t}{RC}})}$
Energy stored in capacitor after time t is given by:
E(t)=q2(t)2C$E(t)={\displaystyle \frac{q^2(t)}{2C}}$
E(t)=CVs(1−e−tRC)2C${\displaystyle E(t)={\displaystyle \frac{{\displaystyle \frac{C}{V_s}}(1-e^{-\frac{t}{RC}})}{2C}}}$

Formula

Time to reach specific percentage of charge in a RC circuit

qq0=1−e−tτ${\displaystyle \frac{q}{q_0}}=1-e^{-{\displaystyle \frac{t}{\tau }}}$
ln(1−qq0)=−tτ$ln(1-{\displaystyle \frac{q}{q_0}})=-{\displaystyle \frac{t}{\tau }}$
t=τln(11−qq0)$t=\tau ln({\displaystyle \frac{1}{1-{\displaystyle \frac{q}{q_0}}}})$

Definition

Sudden change in voltage s forced across a capacitor

If a source of voltage is suddenly applied to an uncharged capacitor, the capacitor will draw current from that source, until the capacitors voltage equals that of the source. Once the capacitor voltage reached this charged state, its current decays to zero. 

Example

Solving complicated RC circuit excited by DC

Example:
Consider the circuit shown in the figure. The capacitor is initially uncharged. Find the current across R2$R_2$ after time t.
Solution:
To find the time-constant, the emf source is short-circuited. The equivalent resistance is then given by:
Req=(R1+R3)||R2$R_{eq}=(R_1+R_3)||R_2$
Req=500011Ω$R_{eq}={\displaystyle \frac{5000}{11}}\mathrm{\Omega }$
Time-constant is then given by: τ=ReqC=122s$\tau =R_{eq}C={\displaystyle \frac{1}{22}}s$
Now, voltage across the capacitor initially is V(0)=0$V(0)=0$ and remains the same at the time of switching.
This is the same as the voltage across R2$R_2$.
By ohm's law, i2(0)=0$i_2(0)=0$
In steady state, the capacitor is fully charged and acts as open circuit. Then current across R2$R_2$ can be found using KCL. 
i2(∞)=20R1+R2+R3=1275A$i_2(\mathrm{\infty })={\displaystyle \frac{20}{R_1+R_2+R_3}}={\displaystyle \frac{1}{275}}A$  
Using general solution,
i2(t)=i2(∞)−[i2(∞)−i2(0)]e−tτ$i_2(t)=i_2(\mathrm{\infty })-[i_2(\mathrm{\infty })-i_2(0)]e^{-\frac{t}{\tau }}$ 
i2(t)=1275(1−e−22t)A$i_2(t)={\displaystyle \frac{1}{275}}(1-e^{-22t})A$ 
Note:
This approach is useful when the resistors and capacitors network can be separated into two groups.

Formula

Heat generated in resistor in RC circuit

A capacitor C$C$ is charged to V0$V_0$ volts, a resistor R$R$ is connected across it. The heat produced in the resistor is CV20/2$C{V}_0^2/2$

Definition

Direct Current

Direct Current- flows has constant magnitude and flows in particular direction.
A Battery is good example of DC power source.

Example

Sources of direct current

Direct current is a current of constant magnitude flowing in one direction. It is generally produced by electrochemical cells in which chemical energy is converted into electrical energy. A cell consists of electrodes and electrolyte. 

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