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In the given figure,we connect three cells of different emf which are $V_1,V_2,V_3$ with a resistance $R$ in the first loop. Now suppose we emit current $I$ from $V_1$. $I$ gets divided into $I_1$ and $I_2$ with $I_1$ flowing through $V_2$ and $I_2$ through $V_3$.

Now if i apply KVL in the second loop where their are only voltage sources $V_2$ and $V_3$, we will get $-V_3+V_2=0$ or $V_2=V_3$ but that's a contradiction since we took the $3$ cells of different emfs. Where am i making a mistake? Is KVL not applicable to this loop? Or did i make a mistake in dividing the currents? Kindly correct me.

ADDENUM: As per the answers,i came to the conclusion that it is impossible to make such a circuit. But the following is an old problem from an olympiad.(A previous post of mine was closed alleging it to be a homework problem even though i was making sure if the problem was wrong or not)

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All the diodes are ideal here so that should mean there is no potential drop and they are just behaving as a normal wire which is the situation i have in the original question,then how is the circuit they made even possible to exist? As per the above answers,if my given circuit cant exist,then so can't the circuit in this problem.

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    $\begingroup$ No source is ideal in real life. They all have output (series) resistance so there are no contradictions in real life. Ideal things simplify analysis but can't truly exist because somewhere there is a scenario where a contradiction that will arise if they did. What you're doing exactly like a student setting up a problem with an unstoppable force meeting an immovable object and then asking what went wrong. You need to know the limits of your model. $\endgroup$
    – DKNguyen
    May 16, 2022 at 0:24
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    $\begingroup$ A diode with Vf=0V ist just a normal wire because it only conducts one way. That diode circuit is fine because only one diode will conduct (the one with lowest voltage source producing the greatest voltage difference across the diode, and when it does the other diodes become reverse biased. $\endgroup$
    – DKNguyen
    May 16, 2022 at 2:25
  • $\begingroup$ Could you please explain a bit elaborately? Sorry for asking too much but i am still a beginner. $\endgroup$
    – madness
    May 16, 2022 at 2:49
  • $\begingroup$ what is the emf for the first diode. You blocked it with the cursor. $\endgroup$
    – Bob D
    May 16, 2022 at 2:49
  • $\begingroup$ That's 1 volt.. $\endgroup$
    – madness
    May 16, 2022 at 2:49

2 Answers 2

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The mistake you are making is trying to connect two ideal batteries $V_2$ and $V_3$ with different EMFs in parallel, winding up with the inconsistency of having two different voltages in parallel, whereas the voltage across any two parallel components has to be the same.

When connecting batteries in parallel, you need to include the internal resistance that exists in all real batteries.

Your ADDENDUM:

There is no inconsistency here because only the diode with the 1 volt emf battery will be forward biased and conduct. Then the voltage across that diode/emf combination will be 1 volt with a voltage drop across the 1 K resistor being 9 volts. The one volt drop will keep the other two diodes reverse biased.

One might ask, "why is it assumed only the first diode conducts given the fact that the voltage of the supply battery exceeds all of the cathode voltages?". The answer would be, assuming all the wires connecting the anodes have zero resistance, there can only be one voltage common to all the anodes.

Let's assume that the second diode (from the left) conducts instead of the first. That would make the anode potential for all three diodes 4V. The third diode will still be reverse biased and therefore not conduct. But the first diode will be forward biased and conduct. If that actually happened, then we would simultaneously have two different potentials, 1V and 4V, for the same electrical point in the circuit.

Hope this helps.

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This entire issue can be avoided given that we do not use the idealisation that the potential difference across an ideal wire is zero. This idealisation is not true. The presence of some "internal resistance" can be used to fix this, but a better way is to just accept there IS potential difference across an ideal wire.

Analysing the second loop, can be done by imagining there are 2 different loops, one with one battery and one with another.

Potential is a scalar and so can then be added up to find the PD across a closed loop.

The first loop starting anticlockwise from the positive terminal is

$$- V_{2} + V_{2} = 0$$

The second

$$ -V_{3} + V_{3} = 0$$

The total PD from both batteries is

$$- V_{2} + V_{2} + -V_{3} + V_{3} = 0$$

There is no error that the potentials are the same, the only reason you got that result is by ignoring the PD across the wires.

More directly, this can also fix the apparent contradiction that the potential from each branch is the same, despite having batteries of different potentials. Potential decreases along the wire to make up for this difference, it is not just the batteries EMF inside the battery you need to worry about.

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