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New to electronics. Can't get my head around so many basic things but the application of KVL is bugging me the most.

I want to know what happens when Kirchoff's voltage law is broken. If for example, I have a 6V battery and a 4V load in a closed circuit. What happens here? The 4V load will consume or result in a 4V voltage drop. There will be a 2V "spare" unused voltage remaining before the current hits the negative terminal of the battery. In this instance, KVL's law is broken (well, I know it can't be).

I think the right way would be to put a resistor with a 2V voltage drop before the current hits the 4V load? That way KVL is not broken.

Which brings me to my next point: how do I know how much voltage drop a certain resistor will cause? If I used a 10 ohm resistor as a load on a 30V power supply, it will draw 3 amp current. Would this resistor cause a 30V voltage drop? If so, would that mean if I put two 10 ohm resistors on the same circuit (instead of one), each resistor will cause a 15V voltage drop?

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    \$\begingroup\$ What happens when KVL is broken? Exactly this: s2.favim.com/orig/28/… \$\endgroup\$ – Majenko Oct 11 '14 at 23:42
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    \$\begingroup\$ Unfortunately for your theory, there's no such thing as a "4 volt load". If your source applies 6v then your load has to deal with 6v, whether it likes it or not. KVL is not so much a law (which may or may not be breakable) as it is an observation of physical facts. \$\endgroup\$ – brhans Oct 12 '14 at 1:41
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To me, Kirchoff's Laws are scientific wording of what should be common-sense observations of electric circuits. Unfortunately, common sense isn't as common as we might like.

KVL says that the total of the voltage drops in a circuit must equal the supplied voltage. If that was not true, we would have some voltage across a wire which would result in approximately infinite current in that wire - but KCL insists that the current is the same at all points in a simple series circuit, so we can't have a huge current at one point in the circuit.

If you connect a device that normally requires 4 volts across a 6 volt battery, sufficient current will flow to make the resulting circuit comply with KVL. The voltage across the "4 volt device" will rise, and the output voltage of the battery will fall (due to a voltage drop across the internal resistance of the battery) such that the voltage across the battery and the device are equal. This may result in the destruction of the 4 volt device, if it cannot withstand the extra voltage and current.

Regarding your second point: the voltage drop across a resistor will depend on its resistance, and on the current passing through it, in accordance with Ohm's Law.

For your example of a 10 Ohm resistor across a 30 volt supply, the current through the resistor will be 3 amps, and the voltage across the resistor will be 30 volts. If you add a second 10 Ohm resistor in series, the load on the power supply is now 20 Ohms, so, by Ohm's Law, the current through the resistors will be 1.5 amps, and there will be 15 volts dropped across each resistor, for a total voltage drop of 30 volts.

If you put the two 10 Ohm resistors in parallel across the 30 volt supply, each resistor will now see 30 volts, and will each pass 3 Amps, so the supply will have to supply 6 Amps.

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  • \$\begingroup\$ Hey thanks so much for the good answer. Really helpes. id upvote your answer but dont have enough rep points. \$\endgroup\$ – Gil Oct 12 '14 at 3:57
  • \$\begingroup\$ @HoGil You should be able to now. :) \$\endgroup\$ – JYelton Oct 13 '14 at 16:57
  • \$\begingroup\$ @JYelton Finally! \$\endgroup\$ – Gil Oct 13 '14 at 19:06

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