0
$\begingroup$

I know this is a very noob question, but i just can't help it. Anyways,here is pretty simple circuit

Now after the circuit is changed this way, why is the current going to be infinitely high?? I hope the question is clear.

$\endgroup$
1
$\begingroup$

I am no expert but I will try to give you reason for your question..

As Ohm's law states that the current $I$ through a conductor between two points is directly proportional to the voltage $V$ across the two points with proportionality constant, resistance $R$

Source : Wikipedia

i.e $$I =\frac{V}{R}$$

in which if voltage $V$ is constant, current $I$ will be inversely proportional to resistance $R$ so if resistance decreases, current will increase i.e $$I \propto \frac{1}{R}$$

So in your first diagram, putting value in equation we get :

$$50.10^{-3} =\frac{6}{R} => R = 120 \Omega$$

And in your second diagram, when switch is closed circuit will be shorted so circuit resistance $R$ will approach $0$ (R~0)

$$I =\frac{6}{0} $$

Current $I$ will approch Infinite ($I$ ~ undefined)

$\endgroup$
  • $\begingroup$ Oooh, this makes sense! But what would make us say that the resistance would be almost zero? Like why not 60 or 30 or 50 etc. $\endgroup$ – Abdel Rahman Shamel Dec 9 '15 at 13:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.