Electromagnetic Induction Concept Page - 8

Example

RLC circuits excited by DC during switching and steady state

Example:
Consider the given circuit with initially uncharged capacitor and inductor. The switch is closed at time t=0. Find the value of current in the circuit and charge across capacitor during switching and in steady state.
Solution:
During switching,
   Current across inductor and voltage across capacitor does not change suddenly. 
   i(0)=0$i(0)=0$
   q(0)=0$q(0)=0$
Now, current flows and capacitor starts charging. The charging occurs until the capacitor potential equals the emf of the voltage source and beyond this point charging stops.
At steady state,
   i(∞)=0$i(\mathrm{\infty })=0$ as capacitor acts as open circuit
   q(∞)=CV$q(\mathrm{\infty })=CV$ 

Definition

Wheatstone bridge in inductors

The attached figure shows the figure of a wheatstone bridge of inductors. In balance condition, VB=VD$V_B=V_D$ and no current flows through L5$L_5$ and hence it acts as an open circuit.
The condition for achieving this balance is L1L4=L2L3$L_1{L}_4=L_2{L}_3$

Example

Problems on inductor circuits

Example: In the given circuit (fig), the switch is closed at t=0$t=0$. Find the current in the inductor when the circuit reaches the steady state and the net change in flux in the inductor.

Solution:
When steady state is achieved, all of the current tends to pass through the inductor and it acts as a no resistance wire. Hence current in that part=12V3Ω=4A.${\displaystyle \frac{12V}{3\mathrm{\Omega }}}=4A.$
Flux when S is closed=LI=0.5×4=2Wb$LI=0.5\times 4=2Wb$
When S was open, current in the inductor was due to 6 volts battery=6V3Ω+3Ω=1A${\displaystyle \frac{6V}{3\mathrm{\Omega }+3\mathrm{\Omega }}}=1A$
Flux when S is open=LI=0.5×1=0.5Wb$=LI=0.5\times 1=0.5Wb$
Thus change in flux=1.5Wb$1.5Wb$

Example

Problems on inductor circuits with varying magnetic field

Example: A closed loop of cross-sectional area 10−2m2${10}^{-2}{m}^2$ which has inductance L=10mH$L=10mH$ and negligible resistance is placed in a time-varying magnetic field. Figure shows the variation of B with time for the interval 4 s. The field is perpendicular to the plane of the loop (given at t=0,B=0,I=0)$t=0,B=0,I=0)$. Find the value of the maximum current induced in the loop.

Solution:
Induced e.m.f. e=LdIdt=AdBdt$e=L\frac{dI}{dt}=A\frac{dB}{dt}$
⇒∫I0dI=∫B0ALdB⇒I=ALB$\Rightarrow {\int }_0^{I}dI={\int }_0^B\frac{A}{L}dB\Rightarrow I=\frac{A}{L}B$
⇒Imax=ALBmax=10−210×10−3×0.1=0.1A=100mA$\Rightarrow I_{max}=\frac{A}{L}{B}_{max}=\frac{{10}^{-2}}{10\times {10}^{-3}}\times 0.1=0.1A=100mA$

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8