Alternating Current Concept Page - 4

Result

General solution of a differential equation for a resistor in LR circuit

Decay of current as the battery is disconnected, the current decreases in the circuit. This  induces an emf (−Ldidt)$(-L\frac{di}{dt})$ in the inductor. As this the only emf in the circuit,

−Ldidt=Ri$-L\frac{di}{dt}=Ri$

Or,  dii=−RLdt$\frac{di}{i}=-\frac{R}{L}dt$.

At t=0,i=i0$t=0,i=i_0$. If the current at time t be i,

∫ii0dii=∫t0−RLdt${\int }_{i_0}^i\frac{di}{i}={\int }_0^t-\frac{R}{L}dt$

Or,  lnii0=−RLt$ln\frac{i}{i_0}=-\frac{R}{L}t$

Or,  i=i0e−tR/L$i=i_0{e}^{-tR/L}$

Or,  i=i0e−t/τ$i=i_0{e}^{-t/\tau }$

Where τ=L/R$\tau =L/R$ is the time constant of the circuit.

We see the current does not suddenly fall to zero. It gradually decreases as time passes. At t=τ$t=\tau$,

i=i0$i=i_0$   k=0.37i0$k=0.37i_0$

Definition

Voltage in a growing LR circuit

In the circuits as shown in the figure, S1$S_1$ and S2$S_2$ are switches. S2$S_2$ remains closed for a long time and S1$S_1$ is opened. Now S1$S_1$ is also closed. Just after S1$S_1$ is closed, find the potential difference (V$V$) across R$R$ and didt${\displaystyle \frac{di}{dt}}$ in L$L$.Before closing S1$S_1$, current in the inductor is i=ε2R$i={\displaystyle \frac{\epsilon }{2R}}$. 

Just after closing S1$S_1$, current in inductor will remain same, and other currents are as shown
For left loop: ε=i1R+L(didt)$\epsilon =i_{1}R+L({\displaystyle \frac{di}{dt}})$ .....(i)
For right loop: ε=i2(2R)+L(didt)$\epsilon =i_2(2R)+L({\displaystyle \frac{di}{dt}})$ .....(ii)
and i1+i2=i$i_1+i_2=i$ .....(iii)
Solve to get didt=2ε3L${\displaystyle \frac{di}{dt}}={\displaystyle \frac{2\epsilon }{3L}}$
p.d. across R$R$: VR=i1R=ε−L(didt)=ε−L(2ε3L)=ε3$V_R=i_{1}R=\epsilon -L({\displaystyle \frac{di}{dt}})=\epsilon -L\left(\frac{2\epsilon }{3L}\right)={\displaystyle \frac{\epsilon }{3}}$

Example

Current in a growing/decaying RL circuit

An emf of 15$15$ V$V$ is applied in a circuit containing 5$5$ H$H$ inductance and 10Ω$10\mathrm{\Omega }$ resistance. The ratio of the currents at time t=∞$t=\mathrm{\infty }$ and t=1$t=1$ s$s$ is:
i=ER(1−e−RtL)$i={\displaystyle \frac{E}{R}}(1-e^{{\displaystyle \frac{-Rt}{L}}})$

It=∞=I0${\displaystyle I_{t=\mathrm{\infty }}=I_0}$

It=1s=I0(1−e−105×1)${\displaystyle I_{t=1s}=I_0(1-e^{-\frac{10}{5}\times 1})}$

=I0(1−e−2)=I0(1−1e2)${\displaystyle =I_0(1-e^{-2})=I_0(1-\frac{1}{e^2})}$

∴=It=∞It=1s=I0e2I0(e2−1)=e2e2−1${\displaystyle \therefore ={\displaystyle \frac{I_{t=\mathrm{\infty }}}{I_{t=1s}}}=\frac{I_0{e}^2}{I_0(e^2-1)}={\displaystyle \frac{e^2}{e^2-1}}}$

Example

Energy stored in an inductor in an LR circuit

A current of 2$2$ A is increasing at the rate of 4$4$ A/s$A/s$ through a coil of inductance 2$2$ H. The energy stored in the inductor per unit time is:Energy stored per unit time is
=dUdt$={\displaystyle \frac{dU}{dt}}$
=ddt(12Li2)$={\displaystyle \frac{d}{dt}}({\displaystyle \frac{1}{2}}L{i}^2)$
=Lididt$=Li{\displaystyle \frac{di}{dt}}$
=2×2×4$=2\times 2\times 4$
=16W$=16W$

Example

Calculate amount of heat generated in resistors in LR circuits

Example: A solenoid of inductance L$L$ with resistance r$r$ is connected in
parallel to a resistance R$R$. A battery of emf E$E$ and of negligible internal resistance is connected across the parallel combination as shown in the figure. At the time t=0$t=0$, switch S$S$ is opened, calculate amount of heat generated in the solenoid.

Solution:
Steady state current developed in the inductor =Er=i0$=\frac{E}{r}=i_0$ (say)
Now this current decrease to zero exponentially through r$r$ and R$R$
i=i0e−t/τL$i=i_0{e}^{-t/{\tau }_L}$
Where τL=LR+r${\tau }_L=\frac{L}{R+r}$
Energy stored in inductor
U0=12Li02$U_0=\frac{1}{2}L{i_0}^2$
(12L)(Er)2$\left(\frac{1}{2}L\right){\left(\frac{E}{r}\right)}^2$
Now this energy dissipates in r$r$ and R$R$ in direct ratio of resistance.
∴Hr=(rR+r)U0$\therefore H_r=\left(\frac{r}{R+r}\right)U_0$
=E2L2r(R+r)$=\frac{E^{2}L}{2r(R+r)}$

Example

Calculate potential difference when the current in the circuit is maximum

Example: Two capacitors of capacitance C and 3C are changed to potential difference V0$V_0$ and 2V0$2V_0$, respectively, and connected to an inductor of inductance L as shown in fig. Initially, the current in the inductor is zero. Now, switch S is closed. Find the potential difference across capacitor of capacitance 3C when the current in the circuit is maximum.

Solution:
The state of the circuit 't' seconds after the switch is closed can be shown as in figure where 'q' is charge as a function of time and 'i' is the current in the circuit.By applying Kirchoff's law, potential drop across any loop is zero.
Therefore,−(CV0+qC)−(6CV0+q3C)−Ldidt=0$-({\displaystyle \frac{C{V}_0+q}{C}})-(\frac{6C{V}_0+q}{3C})-L{\displaystyle \frac{di}{dt}}=0$
−3V0=4q3C+Ldidt$-3V_0={\displaystyle \frac{4q}{3C}}+L{\displaystyle \frac{di}{dt}}$
When current is maximum, didt=0${\displaystyle \frac{di}{dt}}=0$
q=−9CV04$q=-{\displaystyle \frac{9C{V}_0}{4}}$
Potential drop across the 3C capacitor then becomes15CV043C=5V04${\displaystyle \frac{{\displaystyle \frac{15C{V}_0}{4}}}{3C}}={\displaystyle \frac{5V_0}{4}}$

Definition

Current in AC circuit containing resistance and capacitance

Let an AC of EMF E be connected to a series combination of a capacitor of pure capacitor C and resistor R.
The source voltage is given by : V=VoSinωt$V=V_{o}Sin\omega t$  
Voltage across resistor is given by : VR=IR$V_R=IR$
Voltage across Inductance is given by :Vc=IXC$V_c=I{X}_C$
The phase difference between VR$V_R$ and VC$V_C$ will be 90 Degrees
The resultant voltage will be V2=V2R+V2C$V^2=V{}_R^2+V{}_C^2$
                                                       =IR2+X2C−−−−−−−√$=I\sqrt{R^2+X{}_C^2}$ 
Comparing this with ohms law we get impedance Z
Z=R2+1ω2C2−−−−−−−−−√$Z=\sqrt{R^2+{\displaystyle \frac{1}{{\omega }^2{C}^2}}}$
Phase difference will be given by ϕ=tan−11ωCR$\varphi =ta{n}^{-1}{\displaystyle \frac{1}{\omega CR}}$ 
Current is given by : I=Iosin(ωt+ϕ)$I=I_{o}sin(\omega t+\varphi )$ Where ϕ$\varphi$ is phase difference

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