In the diagram I have attached, the top circuit is the schematic equivalent I am trying to use to model the middle circuit which uses real voltaic cells (that have internal resistance). I am not sure if these diagrams are equivalent, specifically whether I can correctly convert the voltaic cells into ideal batteries connected to resistors, because when I performed this experiment, there was enough current to light the bulb in the bottom circuit (the only change from the middle circuit is that the wire containing resistor A was removed) but not in the middle circuit. This contradicts how adding an extra wire in parallel (the wire containing resistor A in this case) drops the equivalent resistance.

enter image description here

  • $\begingroup$ Could you please measure the resistance of the bulb? $\endgroup$ – Yashas May 19 '16 at 5:29
  • $\begingroup$ I actually don't have access to the lab equipment anymore, but could you please explain the effect of the value of the resistance of the bulb? $\endgroup$ – jt2000 May 19 '16 at 6:39
  • $\begingroup$ My answer discusses the effect of the resistance in detail. This question is a physics question. You'd have got an answer much earlier if you had posted the Q in the Physics section. $\endgroup$ – Yashas May 19 '16 at 6:42

The fact that the cell has internal resistance & it acts as a source corrupts your argument. Your argument that connecting a resistance across one of the cells in parallel reduces the overall resistance of the circuit and hence expect the current to increase is wrong.

I will derive an equation to obtain the current & potential drop across the bulb for the middle case (which can be reduced to the bottom case by substituting infinity as the resistance of the resistor that is connected in parallel)

enter image description here

$r$ is the internal resistance of the cells and $E$ is the E.M.F of the cells. Let the current through $R2$ (bulb) be $I$, the current through $R1$ be $i$, the resistance of the resistor $R1$ be $R_1$ and the resistance of the bulb be $R_2$.

Applying KVL to loop the mini-circuit consisting of R1 and the cell which is in parallel to R1.

$$E - (I - i)r + iR_1 = 0$$

Applying KVL to the outer loop consisting of R1, R2, r and E (the one outside the mini-circuit).

$$E - IR_2 - iR_1 - Ir = 0$$

Solving for I, we get

$$I = \frac{E(2R_1 + r)}{2R_1r + R_1R_2 + R_2r + r^2}$$

Power dissipated by the bulb (in other words, the intensity of the light produced by the bulb) is given by $$P = I^2R_2$$

This tells us that larger the current drawn, the brighter your bulb will glow.

You can obtain the current through the bulb for the bottom case by substituting $R_1$ as infinity. Doing so, we get

$$I = \frac{2E}{2r + R_2}$$

which is correct.

Plotting a graph between the current versus resistance of resistor $R_1$, you get

enter image description here

where the red line represents the variation of current with the resistance of $R_1$ (middle case) and the orange line represents the current passing through the bulb if the resistance of $R_1$ was infinity (bottom case).

Note that I have used arbitrary values for the constants which is not a problem since the values only determine the boundary conditions of the graph. The shape will remain same for any set of constants.

You can clearly see that the red line tends to coincide with the orange line as the resistance of $R_1$ approaches infinity which is consistent with our predictions. The orange line is basically an asymptote for the red line.

From the graph, it is clear that by adding a resistance in parallel to one of the cells, the current through the bulb decreases, hence the power and brightness decreases.

Also, note that the power goes as the square of the current. So a decrease of current by half would reduce the power by a factor of 4.

A good fact to be aware of is, "the cell to which a resistor is connected in parallel does not deliver any significant amount of power to the bulb" for low values of resistance $R_1$. The bulb utilizes the energy from the cell which is outside.

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  • $\begingroup$ Actually I made a mistake, the bulb was actually an LED so it should have very low resistance $\endgroup$ – jt2000 May 19 '16 at 6:42
  • $\begingroup$ If you take the resistance of the bulb as 2 ohms. The brightness of the bulb will be just 0.5 times of the brightness of the bulb in the previous case (for low values of the resistance which is connected in parallel). Now given the fact that you used an LED, the LED may not glow at all if it does not recieve the barrier potential drop. A bulb might have given you a bit of glow but an LED won't if the potential drop goes below a cutoff. The answer now makes absolute sense since you used an LED. You can expect the LED to stop glowing completely. $\endgroup$ – Yashas May 19 '16 at 6:48
  • $\begingroup$ Thank you so much for the detailed answer! Looks like I have some reading to do on LEDs because I thought replacing it with a lightbulb would not affect the result in this case... $\endgroup$ – jt2000 May 19 '16 at 6:50
  • $\begingroup$ I actually had one more question. If the cell in parallel to the resistor does not deliver any (is it none at all or just very little?) power to the bulb, where does the cell's power get dissipated? Is it through its internal resistance? $\endgroup$ – jt2000 May 19 '16 at 7:06
  • $\begingroup$ It hardly makes any contribution. Most of it is dissipated in the internal resistance and in the resistor which is connected in parallel. You can use the superposition theorem to find out the contribution of current by the cell to the bulb. It is very tiny. Of course, this is highly dependent on the value of resistance of the resistor. As the resistance gets larger, the more contribution the cell will make...will edit the answer since it kinda says that absolute zero contribution is made by the cell. $\endgroup$ – Yashas May 19 '16 at 7:10

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