# Net Voltage in inductor cicuit

Discussion in 'General Electronics Chat' started by Muhammad AbuBakr, May 20, 2015.

May 20, 2015
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What is the expression for net voltage in a circuit which has a coil connected in series with battery through a rheostat (so current can be changed)?Where in the circuit is back emf present?

2. ### Alec_t AAC Fanatic!

Sep 17, 2013
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The back-emf is generated directly across the coil (i.e between its two terminals) when the current is changed.

3. ### ian field Distinguished Member

Oct 27, 2012
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The back emf is developed across the inductor - but you won't generate much emf by varying a rheostat.

The back emf is produced when lines of magnetic flux surrounding the inductor collapse. This has to happen fairly rapidly to generate much voltage.

May 20, 2015
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I am fine with that.Get this."If the current decreases the back emf will try to aid the battery"By aiding it means the back emf's polarity will be such that it will try to increase the current in the circuit.Am I right?

May 20, 2015
3
0
I am fine with that.Get this."If the current decreases the back emf will try to aid the battery"By aiding it means the back emf's polarity will be such that it will try to increase the current in the circuit.Am I right?

6. ### ian field Distinguished Member

Oct 27, 2012
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The action of an inductor is opposite to that of a capacitor - the capacitor will oppose the voltage change by either charging or discharging, and in so doing taking or giving up current.

If you vary the current through an inductor with a rheostat, the magnetic lines of flux around the inductor will expand or contract accordingly - but you'd probably need laboratory instruments to record that in action.

7. ### crutschow Expert

Mar 14, 2008
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Not quite. The induced voltage will be such as to try to keep the current flowing. It will not increase the current.
An inductor's inductance acts rather like the mass inertia in a mechanical system. A mass resists a change in velocity and the inductor resists a change in current. The larger the mass or the larger the inductor, the greater the resistance to change.

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8. ### MikeML AAC Fanatic!

Oct 2, 2009
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The switch closes at 2s and opens at 12s.

In the steady-state with the switch open, the current through the inductor is V1/R2 = 10/2 = 5A. With switch closed, it is 10/(R1//R2) = 10/1 = 10A. Now look at what V(out) must do to satisfy Kirchoff's current law as the switch is closed, then opened. V(out) drops when the switch is closed, but spikes when the switch is opened. The magnitude of the spikes is absolutely determined by what the inductor current was just before the switch transition.