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In an induction coil, When I supply a constant DC voltage of $10 V$ the current goes increasing with time.

Why is this phenomenon taking place in an inductor ?

How much is the lag between the supplied voltage and the back emf produced in an inductor ?

How is this helpful in eliminating fluctuations in current in a circuit. I doubt if there is a change in current, it would ruin the circuit before the back emf is produced .

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2 Answers 2

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The back emf is created by changing current, not steady state current, and is created instantly when the current changes. When you put $10V$ across the coil with its $L$ and $R$, the current starts increasing. The $10V$ gets split between the $L\frac {di}{dt}$ and the $iR$, which is why there is an exponential approach to steady state. At steady state, $\frac {di}{dt}=0$ and the current is $\frac {10}R$. If you are using this to reduce current fluctuations, as soon as the current is reduced, the "back emf" adds to the supply voltage to try and maintain the current. If the current is increased, the back emf reduces the voltage to try and maintain the current.

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The base equation for a RL circuit is:

$$U=Ri+L\frac{di}{dt}$$

If you solve the diff. equation(assuming $i(0)=0$) you get:

$$i(t)=\frac{U}{R}(1-e^\frac{t}{T})$$ $$ T=\frac{L}{R}-\text{This is the time required to get 63% of}\frac{U}{R}$$

Let's assume you already have a current $i(0)=i_1<\frac{U}{R}$, the solution will be:

$$i(t)=i_1+\left(\frac{U}{R}-i_1\right)(1-e^\frac{t}{T})$$

Change the sign of $i_1$ accordingly for other cases. You can calculate what exactly happens each time with these formulas.

Insert your parameters and see what happens. To get $u_L(t)$(or any other voltage) just insert your current in the right equation.


Okay now we have a framework to work with. Let's see what happens!

In the start you have $R$ and $L$. If you add another component let's call it $R_1$ the following will happen:

  1. You start from a zero state $i(0)=0$; and turn the new circuit on.
  2. We get a new current $i(t)=\frac{U}{R+R_1}(1-e^\frac{t}{T})$
  3. The voltage drop across $R_1$ is $u_{R_1}=R_1i(t)=R_1\frac{U}{R+R_1}(1-e^\frac{t}{T})$
  4. The final current through $R_1$ is $i=\frac{U}{R+R_1}$; and the final voltage is $u_{R_1}=R_1\frac{U}{R+R_1}$
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