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I understand both the current divider formula and why it is mathematically correct. I am curious though why physically this is true. My book provided me the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The explanation in the book left a lot to be desired. The physical explanation was basically:

  1. The total current from the source I is split into two currents I1 and I2 at the junction of the parallel resistors. I1 passes through R1 and I2 passes through R2 respectively.

  2. They are recombined at the other side so that the current and voltage across them are made whole again.

From the formula, the current x across a given resistor x is given by:

Ix = It * (Rt)/(Rx + Rt)

and so from the formula it's clear the total resistance (and therefore the other resistors including the branch's resistors) has an effect on a given individual branch of current.

Why physically is this effect true? It seems counter-intuitive in the sense that you might expect only the branch's resistor, rather than both resistors, to effect the current.

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    \$\begingroup\$ Rt is not total resistance here. Perhaps you should choose a different subscript. \$\endgroup\$ – Scott Seidman Mar 4 at 19:52
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    \$\begingroup\$ That formula holds if your circuit is being driven by a current source. The one you've drawn has a voltage source - and your intuition in this case is correct. \$\endgroup\$ – brhans Mar 4 at 19:54
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    \$\begingroup\$ Consider the current per resistor is defined by the voltage that is across each resistor. Then the current through each resistor is independent of the other. \$\endgroup\$ – Huisman Mar 4 at 19:54
  • \$\begingroup\$ I had taken the formula from wikipedia out of sheer laziness @ScottSeidman (en.wikipedia.org/wiki/Current_divider). It seems Rt is indeed total resistance. \$\endgroup\$ – CL40 Mar 4 at 21:37
  • \$\begingroup\$ No - Rt is not the "total" resistance. In the given formula, the current It must be replaced by the total current I=V1/(R1+R2) - that`s all. \$\endgroup\$ – LvW Mar 4 at 21:46
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You are right in assuming that the current through R1 does not depend on the resistor R2 (resp. the current through R2).

However, because both independent currents are driven by a commom voltage source - and because I1+I2=I - you can replace the Voltage V1 in the expression I1=V1/R1 by the other current I2 and I. The same applies vice versa to I2.

By doing this, you arrive at the given expression for Ix (If x=1, we have t=2 and if x=2, we have t=1).

In detail: I1=I[R2/(R1+R2)] and I2=I[R1/(R1+R2)]

Summary: Answering your question (headline), it seems that the resistor R2 would have an influence on the current I1 - however, this is not true. We only have expressed the driving voltage V1 by the total current which also depends on R2 (and vice versa).

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answers. \$\endgroup\$ – Dave Tweed Mar 5 at 17:30
  • \$\begingroup\$ Univ. of Berkeley: "Thus, current entering a set of parallel resistors splits such that more current passes through the less resistive path". You disagree - well, WHAT DO YOU DISAGREE WITH? With the applicability of the mentioned rule for each resistive parallel combination? Or just with the name? I think, the name "current division" for the circuit under discussion is in full accordance with all relevant textbooks. \$\endgroup\$ – LvW Mar 5 at 17:31
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I think the formula is wrong:

If \$I_t\$ is defined as \$I_x+I_y\$ and \$R_t\$ is defined as \$\frac{R_x R_y}{(R_x +R_y)} \$ then

$$ V = I_t * R_t $$ $$ I_x = V/R_x $$

$$ I_x = I_t * R_t/R_x $$ not $$ I_t * (R_t)/(R_x + R_t)$$

If \$I_t\$ is defined as \$I_x+I_y\$ and \$R_t\$ is defined as \$ (R_x +R_y) \$ then the formula given by OP becomes

$$I_x = I_t * (R_t)/(R_x + R_t)$$ $$I_x = I_t * (R_x +R_y)/(R_x + R_x +R_y)$$ which makes no sense either.

Were V2 a current source, then it still makes no sense.
The current source equals \$I_t\$ and V in above equations is still the voltage across both resistors.

So, either it is $$I x =I_t ∗R_t /R_x$$ or like @ScottSeidman already remarked, Rt is not the total resistance.

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  • \$\begingroup\$ x and t are just indices.....and in the original formula, It is wrong and must be replaced by the total current I. \$\endgroup\$ – LvW Mar 4 at 21:48
  • \$\begingroup\$ Like I already did. The first approach equals your answer. \$\endgroup\$ – Huisman Mar 4 at 22:06

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