Understanding Self-Induction: Calculating Current Growth Rate

In summary, the speaker discusses a question about self induction involving a coil with a resistance of 15 ohm, an inductance of 0.6 Henry, and a direct current source of 120 volts. To calculate the rate of growth of the current when it reaches 80% of its maximum value, the speaker first determines the maximum current of 8 Ampere and then calculates the induced emf in the coil to be 96 volts. However, in the book, the answer is found by using the difference in volts between the battery and the induced emf, 24 volts. This is because of Lenz' Law, which states that the induced emf opposes the change in current. Therefore, the difference in voltage must
  • #1
sagda_m2
1
0
This question is about the self induction:

The resistance of a coil is 15 ohm, itself induction is 0.6 Henry, it is connected to a source of direct current which gives 120 volts. Calculate the rate of growth of the current at the moment when the current reaches 80 % of its maximum value.

My Answer:

First I calculated the maximum current:
I= V/R = 120/5 =8 Ampere

Therefore: the 80 % of the maximum current =6.4 A

Then the electromotive force will be:
V= I * R = 6.4 * 15 = 96 volt

And since: the induced e.m.f in the coil = -L * (dI/dt)

Therefore: 96 = 0.6 * (dI/dt)

Therefore: dI/dt = 96/0.6 = 160 A/s

But…..
I found the answer in my book as follows:

Since e.m.f = -L*(dI/dt)

Therefore: (120-96) = 0.6 * (dI/dt)

Therefore: dI/dt = 24/0.6 = 40 A/s

So, why did he substitute the e.m.f in the equation with the difference in volts (120-96)? Why didn't he substitute with the (96 volts)?


Thanks in advance,
 
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  • #2
Consider the emf produced by the coil and the battery as separate and think about Lenz' Law.
 
  • #3


I would say that both approaches are correct and valid ways to calculate the rate of growth of the current in this scenario. The difference in approach may be due to different interpretations of the problem or different assumptions made. In the first approach, the e.m.f is calculated based on the 80% of the maximum current, while in the second approach, it is calculated based on the difference in voltage between the source and the induced e.m.f. Both methods can give accurate results, but it is important to clearly state the assumptions and interpretations made in the calculation. It is also worth considering the context in which the problem is presented and any specific instructions or guidelines given. Ultimately, the important thing is to understand the concept of self-induction and how to calculate the rate of current growth in such scenarios.
 

Related to Understanding Self-Induction: Calculating Current Growth Rate

1. What is self-induction?

Self-induction is a phenomenon that occurs in electrical circuits where the changing magnetic field produced by the current in the circuit induces a voltage in the same circuit.

2. How does self-induction affect current growth rate?

Self-induction can slow down the current growth rate in a circuit because the induced voltage creates a back EMF that opposes the growth of current.

3. What is the equation for calculating the current growth rate in a circuit with self-induction?

The equation for calculating the current growth rate in a circuit with self-induction is dI/dt = (V - L*dI/dt)/R, where dI/dt is the current growth rate, V is the applied voltage, L is the self-inductance of the circuit, and R is the resistance of the circuit.

4. How do you calculate the self-inductance of a circuit?

The self-inductance of a circuit can be calculated by dividing the magnetic flux through the circuit by the current flowing through the circuit, or by using the equation L = N*dΦ/dI, where N is the number of turns in the circuit and dΦ/dI is the rate of change of magnetic flux with respect to current.

5. Why is it important to understand self-induction when working with electrical circuits?

It is important to understand self-induction because it can affect the behavior of circuits, particularly in AC circuits, and can lead to unexpected results if not taken into consideration. It also has practical applications, such as in the design of inductors and transformers.

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