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Sir the ans to this quesn is i3=-5/6 A so this means curretn flows from A to D (opp to current indicated in diagram) but through KVL we found PE at D is always greater than A, C,E hence it will always flow from D to other points. 1) How it is voilating Ienter=Iexit. 2)How current flows from A to D even though D is high PE?
Asked by aaryamanmodern | 22 Jun, 2023, 03:10: PM
Figure shows the given circuit with assumed current distribution.

If we apply Kirchoff's voltage law to the loop BAFEB, we get

from above expression, we get i2 = (1/2) A

If we apply Kirchoff's volatge law to the loop BCDEB , we get

From above expression , we get i3 = (1/3) A

If we apply Kirchoff's current law at junction B , we get

-------------------------------

Let E be the reference point to get potential at different points of circuit.

At point B we have 6 V potential with respect to E . At point A we have -8V potential withrespect to E .
Hence B is at higher potential and E is lower potential , hence we get current i2 from B to A.

Potential of point C is -12V with respect to E . Hence current i3 is passing from higher potential B to lower potential C.

Current ii is current passing through battery of EMF 6V . in a circuit, current will statr from higher potential of battery to lower potential of battery .

But inside the battery, current will pass from lower potential to higher potential .

EMF of battery , i.e., electro motive force of battery drives the current from lower  potential to higher potential.

Chemical energy inside the battery gives the electromotive force to drive the current from lower potential to higher potential.

Hence every thing is perfect. No violation of any law.
Answered by Thiyagarajan K | 23 Jun, 2023, 04:44: AM

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