An inductor of 3.3mH with a series resistance of 12.5 ohms is connected to a 5V dc supply. When the supply is switched off, the circuit current decay to zero in 60 microseconds. What is the value of back e.m.f. generated?
-22V
Inductor Back EMF Calculation
This problem involves calculating the back electromotive force (EMF) generated in an inductor when the current flowing through it decays to zero after the DC supply is switched off. Understanding the behavior of inductors in DC circuits and the concept of induced EMF is crucial here.
Initial Current Determination
When the 5V DC supply is connected to the inductor-resistance series circuit for a long time, the inductor acts like a short circuit because it offers no resistance to a steady (DC) current. Therefore, the initial current flowing through the circuit, before the supply is switched off, can be calculated using Ohm's Law:
\(I_0 = \frac{V}{R}\)
Where:
- \(I_0\) is the initial current.
- \(V\) is the DC supply voltage = 5V.
- \(R\) is the series resistance = 12.5 ohms.
Substituting the given values:
\(I_0 = \frac{5 \text{ V}}{12.5 \text{ } \Omega} = 0.4 \text{ A}\)
Rate of Current Decay
The problem states that the circuit current decays from its initial value to zero in 60 microseconds. This allows us to calculate the average rate of change of current (\(\frac{dI}{dt}\)).
- Initial current (\(I_{initial}\)) = \(I_0 = 0.4 \text{ A}\)
- Final current (\(I_{final}\)) = 0 A
- Time taken for decay (\(\Delta t\)) = 60 microseconds = \(60 \times 10^{-6}\) seconds
The change in current (\(\Delta I\)) is \(I_{final} - I_{initial} = 0 \text{ A} - 0.4 \text{ A} = -0.4 \text{ A}\).
The rate of change of current is:
\(\frac{dI}{dt} = \frac{\Delta I}{\Delta t}\)
Substituting the values:
\(\frac{dI}{dt} = \frac{-0.4 \text{ A}}{60 \times 10^{-6} \text{ s}}\)
\(\frac{dI}{dt} = \frac{-0.4 \times 10^6}{60} \text{ A/s}\)
\(\frac{dI}{dt} = \frac{-400000}{60} \text{ A/s}\)
\(\frac{dI}{dt} \approx -6666.67 \text{ A/s}\)
Back EMF Calculation
The back electromotive force (EMF) generated across an inductor is given by the formula:
\(E_{back} = L \frac{dI}{dt}\)
Here, \(E_{back}\) represents the voltage across the inductor. While Lenz's Law typically introduces a negative sign (\(E = -L \frac{dI}{dt}\)) to emphasize opposition to the change in current, in circuit analysis, the voltage across an inductor terminals \(V_L\) is often given by \(L \frac{dI}{dt}\). Since the current is decaying (\(\frac{dI}{dt}\) is negative), the voltage across the inductor will also be negative, indicating its polarity in relation to the initial current direction.
Given:
- Inductance (L) = 3.3 mH = \(3.3 \times 10^{-3}\) H
- Rate of change of current (\(\frac{dI}{dt}\)) = -6666.67 A/s
Substituting these values into the formula:
\(E_{back} = (3.3 \times 10^{-3} \text{ H}) \times (-6666.67 \text{ A/s})\)
\(E_{back} = -21.99999...\text{ V}\)
\(E_{back} \approx -22 \text{ V}\)
Final Back EMF Value
The calculated back e.m.f. generated in the inductor when the current decays is approximately -22V.
| Parameter | Symbol | Value |
|---|---|---|
| Inductance | L | 3.3 mH (\(3.3 \times 10^{-3}\) H) |
| Series Resistance | R | 12.5 \(\Omega\) |
| DC Supply Voltage | V | 5 V |
| Time for Current Decay | \(\Delta t\) | 60 \(\mu\)s (\(60 \times 10^{-6}\) s) |
| Initial Current | \(I_0\) | 0.4 A |
| Rate of Current Change | \(\frac{dI}{dt}\) | -6666.67 A/s |
| Back e.m.f. generated | \(E_{back}\) | -22 V |
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