# Kirchhoff's Voltage Law

1. Feb 7, 2016

Consider the circuit from minute 6:57:

So I have an electric field there everywhere in the loop and I can write integral of E*dl as being v and I end up with kirchhoff's rule. (you don't have to watch all the video to respond to this question, just the circuit)

But what if I had an ideal current source in series with a resistor? How do we write integral of E*dl in a closed loop in this case? Do we even have an electric field inside the current source?

If not, how do we reconcile the current source with kvl?

Last edited: Feb 7, 2016
2. Feb 8, 2016

### cnh1995

Switching arrangement for a current source is different from that for a voltage source. In a voltage source, switch is placed in series with the source and is "closed" to bring the voltage source into the circuit.

In case of current source, the switch is put "across" the current source and not in series. Also, when the switch is "closed", current source is not supplying current to the external circuit (it is shorted). To bring it into the circuit, you must "open" the switch.

If you place a current source in series with an inductor, the inductor voltage will be infinite. When you place a capacitor across an ideal voltage source, the capacitor current is infinite. There has to be some resistance involved in both the cases(parallel with the inductor and in series with the capacitor) to write KVL and for the currents and voltages to be theoretically finite (practically, they are always finite because there is always some resistance involved).
Hope this helps!

Last edited: Feb 8, 2016
3. Feb 8, 2016

Thanks for answer but I was not referring to replacing the voltage source in THAT circuit with a current source.

Take a separate current source. What is integral of E*dl over it? Does it have an electric field inside like the voltage source?

4. Feb 8, 2016

### cnh1995

You mean a simple current source in series with a resistor(no L and C)?

5. Feb 8, 2016

### cnh1995

In that case, there is definitely an electric field inside the current source and hence, there is some voltage across the current source. To maintain a constant current, resistance of the current source changes with the load, thus changing the voltage across it. V and R both change such that the current I=V/R remains constant.
e.g.Collector current of a bjt opreated in active region.

6. Feb 8, 2016

### CWatters

Yes but it's not fixed. Just as the current through a voltage source isn't fixed.. Compare....

1) A Voltage source...

The voltage is constant.
The current depends on the external circuit.

Lets say you have a voltage source connected to a resistor. The voltage depends on the voltage source, lets call it V. The current will depend on the resistor according to I = V/R

2) A current source...

The current is constant.
The voltage depends on the external circuit.

Lets say you have a current source connected to a resistor. The current depend on the current source, lets call it I. The voltage will depend on the resistor according to V=IR

7. Feb 8, 2016

Ok. I understand. Thanks for answers.