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What is the current through resistance of 6 ohm in the figure. please explain through Kirchoff laws and keep it simple on college level image

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closed as off-topic by John Rennie, ACuriousMind Feb 16 '17 at 12:56

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  • $\begingroup$ This will all depend on what the batteries will do when connected like this. You should not connect a 3V and a 9V battery in parallel. It is possible the 3V battery will die in some way. What is the internal resistance of the batteries $\endgroup$ – hdhondt Feb 16 '17 at 11:16
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    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Feb 16 '17 at 11:44
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In drawing a diagram like that, you implicitly assume idealised electrical elements. For example, the "9V battery" on the right symbolises an ideal voltage source: Its terminals are maintained at a potential difference of $9\;\text{V}$, and the device churns out whatever current is necessary to keep the voltage. The lines represent ideal conductors, i.e. "wires" that can conduct any current without producing a voltage along them.

For many applications, these idealisations are pretty good approximations, but sometimes they are not. A real battery's voltage depends on the current it supplied (modelled in the next approximation by an internal resistance), a wire has a resistance (and capacitance and inductance) etc., so if e.g. currents become "large" you have to account for these effects, because the idealised approximation will give wrong results. (What is large depends on the situation.)

In your case, it's even worse: For ideal voltage sources, the arrangement cannot work. The reason is that the two batteries try to enforce different voltages along the resistor. What really happens depends very much on the voltage sources -- for example, the $9\;\text{V}$ battery might "charge" the $3\;\text{V}$ one until the smaller one breaks. Or something else will happen. The current through the resistor can be anything, it is not determined by your diagram (it's not necessarily zero as a previous version of another answer suggested).


Technical addendum:

For Kirchhoff's laws you need to have some relation between currents and voltages in your diagram, such as $U=R*I$ for the resistor and [$U=9\;\text{V}$, $I=\text{arbitrary}$] for an ideal voltage source. However, these relations are inconsistent for your circuit, so you would need to update them with some more realistic $U=U(I,\text{possibly more parameters})$ relations for your voltage sources.

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  • $\begingroup$ Can you point out the mistake in my answer.. $\endgroup$ – Mitchell Feb 16 '17 at 12:12
  • $\begingroup$ The internal battery's resistances are keys here. Actually this will say how fast 3V will explode. Anyway - this is homework question or no attempt question. $\endgroup$ – jaromrax Feb 16 '17 at 12:16
  • $\begingroup$ @bhavya-sharma: You first get two contradictory results for the current (correct), but then you conclude that the current is zero. That's incorrect - the contradiction just menas that we cannot determine the current from the given information. $\endgroup$ – Toffomat Feb 16 '17 at 12:30
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I hope this isn't homework. The figure depicts a physically meaningless situation. A simple battery is an ideal voltage source. The voltage across the resistor cannot simultaneously be both 3 Volts and 9 Volts. Make up your mind. If you account for (i.e., "model") the output impedance of the batteries, the problem is trivial. Without it, it is meaningless.

Don't try this at home. Really do not. It could be dangerous.

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Let's assume an arbitrary current direction, current $i$ comes from the $9V$ battery's positive terminal and divides into two currents $x$ and $i-x$ which go into the resistor and the battery of $3V$ respectively.

Apply Kirchoff's law in the lower circuit containing the $9V$ battery and the resistor.

$+9 -6x = 0$ $x = 1.5A$

Now in the upper circuit,

$+3 -6x = 0.5A$

We have reached an impasse here, we are getting two different values of current which is not possible. Therefore, Kirchhoff's law will not work here.

And for the question about the value of current refer to @Toffomat's answer.

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  • $\begingroup$ The current will be whatever the resultant voltage is after the 3 batteries have finished their battle, divided by 6 ohm. It will be 0 only if both batteries expire. $\endgroup$ – hdhondt Feb 16 '17 at 11:32
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    $\begingroup$ It is short circuited. The current wont choose the resistor way. And as you see the Kirchhoff law isnt working here. This kind of circuit will catch fire. $\endgroup$ – Mitchell Feb 16 '17 at 11:42

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