# What is the back voltage generated by this inductor in this practice problem?

The circuit is in a steady state, and the switch is lifted at t=0. In (a), i calculated the current i and the voltage v at the instantaneous moment that the switch is lifted. The current is 2A and the voltage is 12V.

For (b), I need to calculate di/dt at the same instantaneous moment. I can reach that value if I can figure out the back emf (VL) of the conductor. Using Kirchoff's Law on the loop present after the switch is opened, I derive the following equation:

VL - 10(2) - VC = 0


Where VL is the back voltage given off by the inductor, 10(2) is the voltage drop across the two resistors in series, and VC is the voltage across the capacitor. Solving for VL then gives me 32V. It this correct? This seems like an unlikely answer. I believe my signs in my kirchoff equation may be wrong. Can someone check this for me?

• Your approach is correct, but you do have a sign error in your KVL equation. I calculate $v_L(t=0^+)=-8\ \mathrm{V}$. I define the direction of $v_L$ to match the given $i$ with passive sign convention. – The Photon Dec 16 '15 at 1:31
• Why should it match the given i? Isn't the back emf supposed to be opposite of the direction of the current? – Sam D20 Dec 16 '15 at 1:59
• I'm not sure that "back EMF" is the correct term for what happens here. The way to think about this is that the current through the inductor can not change instantly. The inductor voltage will take whatever value is necessary to ensure that the waveform of current through it is continuous in time. – The Photon Dec 16 '15 at 2:02
• If you want to think about this in terms of back EMF, you need to consider that the steady state current (established with the switch closed) is positive. The current transient produced when the switch opens is negative (the current will start to drop toward zero). It's this negative transient current that the back-EMF opposes. – The Photon Dec 16 '15 at 2:05
• Oh i see. I thought the VL that comes from VL = L * di/dt was the back emf generated by the inductor. Thanks for the help. – Sam D20 Dec 16 '15 at 2:14

## 1 Answer

The moment the switch opens, the top leg of the inductor rapidly falls negative to try and keep the current of 2A flowing. If the 4 ohms resistor were not present, a large negative voltage would appear at the top of the inductor and it would be large enough to cause the breakdown of air and a spark would form thus keeping current flowing out of the bottom of the inductor and into the very negative top.

But the 4 ohm resistor is there so no spark forms - there is a conventional discharge path for the current. So now the inductor uses the capacitor and 4ohm path for forcing current.

That initial 2A current flows through the 4 ohm resistor and creates a volt drop of 8 volts. The bottom end of the 4 ohm is attached to a capacitor charged at 12V so, the top of the 4 ohm becomes instantly 4 volts.

Given that the 2A also flows through the 6 ohm resistor creating another volt drop of 12 volts means that the voltage at the top of the coil is 12 volts lower than 4 volts i.e. -8V.