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The above diagram shows a circuit containing two cells. Why doesn't the current produced due to one cell oppose the current produced by the second. In other words, at the junction C, does any current from the cell having voltage V1 branch into the loop containing the cell V2?

Also, why do we take the potential drop across a resistor to be negative if the current through the resistor is opposite to the direction we choose the loop in? What is the theoretical explanation behind it?

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    $\begingroup$ I have added the picture from the link to the question. $\endgroup$ – Steeven Sep 9 '16 at 16:00
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Why doesn't the current produced due to one cell oppose the current produced by the second. In other words, at the junction C, does any current from the cell having voltage V1 branch into the loop containing the cell V2?

If $V_2 >> V_1$ this could indeed happen : current could be driven backwards through battery $V_1$. You can only find out by solving the circuit. Then if you find that $I_1 < 0$ you will know that the current is driven backwards through battery $V_1$.

Also, why do we take the potential drop across a resistor to be negative if the current through the resistor is opposite to the direction we choose the loop in? What is the theoretical explanation behind it?

A "negative potential drop" is a double negative. It is (perhaps?) ambiguous. Does this mean a potential gain, or just a potential drop?

The choice of current direction in your diagram does not matter. The correct direction will be indicated by the sign of the current (when you find out what that is). What matters is that you label potential drops and rises consistently. When current flows through a resistor from A to B, there is always a potential drop from A to B. (In this I agree with Steeven.)

The reason for this is that potential difference is related to change in potential energy. When current flows through a resistor electrical energy is dissipated as thermal energy. The electrical energy available is reduced. Therefore there is a drop in potential.

In your circuit diagram, assuming (as you have done) that current $I_1$ (which I shall call $I_{BC}$) goes from B to C through $R_1$, there is a potential drop across $R_1$ in the direction of the current - as you have indicated : $$V_{BC}= - I_{BC} R_1$$ However, if you find that current $I_{BC} < 0$ so that it actually goes from C to B through $R_1$ then the "potential drop" which you have indicated is reversed : $$V_{BC} = - I_{BC} R_1 = -(-I_{CB}) R_1 = + I_{CB} R_1$$ $$V_{CB} = -V_{BC} = - I_{CB} R_1$$ The last line indicates again that if we measure the PD in the same direction as the current, there is always a potential drop across the resistor.

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  • $\begingroup$ I meant a negative potential drop according to Kirchoff's Voltage Law. Sorry if it came off any other way. Also, I understand your last statement very well. Which is why I'm trying to understand the theoretical reason behind getting a -ve potential drop across a resistor according to KVL SIGN CONVENTIONS. Not an actual potential gain. $\endgroup$ – LeroyJD Sep 9 '16 at 16:46
  • $\begingroup$ Thanks. I have updated my answer since you posted your comment. Please let me know if you are still confused. $\endgroup$ – sammy gerbil Sep 9 '16 at 17:12
  • $\begingroup$ Still a little hazy on the explanation I'm sorry. Could you try explaining it in terms of a Wheatstone? Because that has both a positive and negative drop in any one of the two possible loops, again, according to KVL conventions. $\endgroup$ – LeroyJD Sep 9 '16 at 18:31
  • $\begingroup$ Sorry Leroy I don't understand what it is that you want me to explain. With reference to your Wheatstone diagram, please can you update your question to make it clear what is still confusing you? Perhaps if you demonstrate how you apply KVL sign conventions to the Wheatstone Bridge and what confusing result you get? $\endgroup$ – sammy gerbil Sep 9 '16 at 19:33
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    $\begingroup$ If you assumed that I1 flows from B to C as in the diagram, then you found that I1 is -ve, this means that in fact I1 flows from C to B. $\endgroup$ – sammy gerbil Sep 16 '16 at 22:09
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Why doesn't the current produced due to one cell oppose the current produced by the second?

Why should it oppose it? When the current "meets" the other current, yes they would be opposed and cancel out if they had nowhere to flow. But they do... The current simply passes down through the middle branch.

why do we take the potential drop across a resistor to be negative if the current through the resistor is opposite to the direction we choose the loop in?

The choice of loop direction doesn't matter. You just must be aware that some components reduce the potential (there is a potential drop across them) - like resistors - while other components increase the potential (there is a rise in potential across them) - like batteries and other sources.

When adding them up (they must sum up to zero for a loop) you just call all potential changes positive if they are a rise, and negative if they are a drop. If you chose the opposite loop direction all rises would look like drops and all drops would look like rises. They all therefore get opposite signs in your equation, but since they equal to zero, that doesn't matter - you just must be consistent, and that is why you choose a loop direction in the first place.

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  • $\begingroup$ I understand that irrespective of whether the loop is taken clockwise or anticlockwise, the final answer remains the same. But if you took a wheatstone bridge and tried to derive the condition for current through a galvanometer to be zero, you would have to consider potential drop across some resistors to be zero. What's the theoretical explanation behind that? $\endgroup$ – LeroyJD Sep 9 '16 at 16:07
  • $\begingroup$ "tried to derive the condition for current through a galvanometer" ... I do not follow you. Could you extend your question with this in a new diagram to show the setup clearly? And if the question turns out to be very electronics-based, the Electronics SE site might be a better place to ask. $\endgroup$ – Steeven Sep 9 '16 at 16:09
  • $\begingroup$ @LeroyJD Link doesn't work. $\endgroup$ – Steeven Sep 9 '16 at 16:14
  • $\begingroup$ goforaplusplus.files.wordpress.com/2011/09/circuit.png?w=604 $\endgroup$ – LeroyJD Sep 9 '16 at 16:16
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    $\begingroup$ @LeroyJD No. If resistance R3 is very large, then it prevents current from moving along this path. It will appear as if there was no path at all. $\endgroup$ – Steeven Sep 9 '16 at 17:01

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