# Why in a DC series Circuit do bulb attached in parallel to highest resistance glow brightest? [closed]

I was performing an experiment in which there were five resistances connected in series and then I connected a bulb with each of the resistance. Each bulb was connected in parallel to each resistance and I observed that The bulb with the highest resistance lit brightest.

All the bulbs had the same resistance of 10 ohms and as it was a series circuit definitely there was same current passing through every point on the circuit.

But the current through the largest resistance or the highest resistance was quite ‘Low‘ and at the same time the current passing through the bulb that was connected with the highest resistance was ‘Hi’ so I just thought that it might be because that resistance was high and the resistance of the bulb was low so the current found the easiest path and just passed through there.

But there must be a reason according to some formula that proves this phenomenon.

• A picture is worth 1000 words. – David White Feb 22 '19 at 20:19
• The formula is called Ohm’s law. Such circuits are analyzed using Kirchhoff’s voltage and current laws. – flaudemus Feb 22 '19 at 21:34
• @WaqadArshad: note that we do not provide answers for homework-like questions here. – flaudemus Feb 22 '19 at 21:37
• @flaudemus I know. But my teacher asked to search for it online. That's why I asked it here! – Waqad Arshad Feb 22 '19 at 21:56

First, when you connect a bulb in parallel with the each resistor, you no longer have a series circuit so it isn't true that the current through each circuit element is identical.

Second, the equivalent resistance of each parallel resistor -bulb combination of is

$$R_{i,eq} = 10\,\Omega || R_i = \frac{10\,\Omega \cdot R_i}{10\,\Omega + R_i},\quad 1\le i \le 5$$

Thus, $$R_{i,eq} < 10\,\Omega$$, and the larger the value of $$R_i$$, the closer $$R_{i,eq}$$ is to $$10\,\Omega$$.

Now, by voltage division, the largest voltage is across the largest equivalent resistance and, of course, the bulb with the largest voltage across will glow brightest.

Then, by current division, the larger the value of $$R_i$$, the larger the proportion of current through the bulb.

There is a simple and elegant way to understand why this happens. When two resistances are connected in parallel, say $$R_1$$ and $$R_2$$, their effective resistance is given by $$R= \frac{R_1 R_2}{R_1+R_2}$$. Here, $$R$$ will be smaller than the smallest of the two resistances. $$R$$ will approach the smaller of the two resistances as the other resistance increases. You can check this by calculating the limit when one of the resistances tends to $$\infty$$, which I leave up to you.

Coming back to your question, the resistance of the combination of the highest resistance and the bulb will be the highest in the circuit. Hence, the potential drop across it will be the highest because, from Ohm's law $$V=IR$$ and since $$I$$ is same for all the resistance-bulb combinations, $$V \propto R$$. Now, the power dissipated will be given by $$\frac{V^2}{R_b}$$ where $$R_b$$ is the resistance of the bulb. Since the potential drop across it is the highest, power dissipated by it will be highest and hence, it will be brightest!!