Category: 13 Current of Electricity and DC Circuits

034 Why did bulb Y become dimmer?

For simplicity, let’s assume that the resistances of the bulbs are constant.

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In the original arrangement, by the potential divider principle, VAB = VBC = 2 V. So bulbs X and Y were equally bright.

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When bulb Z was added in parallel to bulb Y, the effective resistance between nodes B and C was halved. By the potential divider principle, the potential difference across bulb X has increased to 2/3 of 4 V = 2.67 V, whereas the pd across bulb Y (and bulb Z) has dropped to 1.33 V. This explains why bulb X now shines brighter than before, whereas bulb Y dims compared to before.

023 Why did the 35 W bulb outshine the 50 W bulb?

A bulb with a 12 V 50 W rating is designed to have a resistance of 122/50 = 2.88 Ω when in operation (P=V2/R). Similarly, a bulb rated at 12 V 35 W rating is designed to have an operating resistance of 122/35 = 4.11 Ω. The point to note is that a 50 W bulb has a smaller resistance than a 35 W bulb.

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At first both bulbs were individually connected across a 6 V battery. Since the potential differences across both bulbs were the same, to compare the power dissipated in the two bulbs, we should think V2/R. So the 50 W bulb, with a smaller resistance, shone brighter.

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Later the bulbs were connected in series across a 12 V battery. Since the current flowing through both bulbs was the same, to compare the power dissipated in the two bulbs, we should think I2R.  So the 35 W bulb, with larger resistance, shone brighter.

(Do note that when connected in series, the potential difference across each bulb was no longer 6 V each. By the potential divider principle, the 35 W bulb, with its larger resistance, ended up with more than 6 V of potential difference across, while the 50 W bulb had less than 6 V across)

015 Why can’t ohmmeters work together?

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For ease of discussion, let’s call the Digital Multimeter (DMM) on the left DMM A, and the one on the right DMM B.

To measure resistance, DMM A actually drives a constant current I0 through its terminals. It then measures the potential difference V0 across its two terminals. Based on V0, the DMM reports the resistance R0.

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What happens when DMM B joins in the fray? Since DMM B will also be driving a constant current I0 through its terminals, the total current will be 2I0,  resulting in a pd of V0=2I0R0 across R0. But both DMMs have no idea of each other’s presence and assumes a current of I0 through R0. They thus (wrongly) reports a resistance of 2V0/I0 = 2R0.

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When DMM B’s range was increased to 2 MΩ, it expects a larger resistance, so it drives a smaller current of 0.1I0 instead. This results in a total current of 1.1I0 through the resistor, resulting in a pd of V0=1.1I0R0. DMM A, which assumes a current of 1I0, reports a (less inaccurate) resistance of 1.1R0. But DMM B, which assumes a current of 0.1 I0, reports (an even more inaccurate) resistance of 11R0.

The moral of the story is, when using the DMM as an ohmmeter, it must be the only EMF source in the circuit. If not, the readings will all be screwed up.

013 Can you solve the Resistor’s Cube?

Resistance across the cube

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Let’s imagine a current I passing across the cube. By symmetry, the current will split equally at each junction. We can thus tell that the current at each of the blue, green and red resistors will be I/3, I/6 and I/3 respectively.

This means that the potential difference across the cube will be

V = I/3 *R + I/6 *R + I/3 R = 5/6 IR.

Since resistance is the ratio of V to I, Reff = 5/6 R 

Resistance across a face

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If the two red resistors were not there, then the effective resistance between points a and b is simply

Rab     = ( 2R || 2R ) || [ R + (2R||2R) + R]

= R || 3R

= ¾ R

Then we notice that points c and d are at the same potential. This means even if the two red resistors were there, no current will pass through them anyway. This means the two red resistors can be removed without affecting the circuit in any way.

Resistance across an edge

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By symmetry, we can tell that points c and d are at the same potential, so these two points can be shorted together without affecting the circuit in any way. Likewise, points e and f can be shorted together since they are at the same potential.

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The circuit can now be redrawn as shown above. The effective resistance between point cd and ef can be simplified to

Rcd,ef = [(R||R) + R + (R||R)] || R || R

= 2R || R || R

= 2/5 R

The effective resistance across an edge will thus be

Rab     = [(R||R) + 2/5 R + (R||R)] || R

= 5/7 R || R

= 7/12 R

 

011 Why is the last bulb so slow?

The real mystery in this video, is why it takes longer to light up the fifth bulb if the other four were already lit up, compared to if all five were lit together.

The explanation has to do with the fact that resistance of a filament increases with temperature. So a bulb that is already lit up has a larger resistance than a cold bulb.

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In this video, when the 5 bulbs are all lit up from cold together, because they grow hot together at the same rate, their resistances are the same at any one time. So the potential difference across each bulb is always 46 V (230 V /5 bulbs).

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When 1 bulb is lit up from cold, with all the other 4 already hot, the initial potential difference across the cold bulb is much less than 46 V. This is because the other 4 hot filaments, with their larger resistances, has more than 46 V across each of them, as the potential divider principle dictates.

The pd across the 5th bulb will however eventually grow to become 46 V, when the bulb has heated up and its resistance becomes the same as all the other bulbs.

Compared to a bulb with 46 V across right from the beginning, a bulb with pd increasing slowly towards 46 V takes a longer time to light up.

007 Bulbs together light up faster? Bulb on its own light up slower?

The resistance of a filament increases with temperature. So a bulb that is already lit up has a larger resistance than a cold bulb.

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In this video, when the 10 bulbs are all lit up from cold together, because they grow hot together at the same rate, their resistances are the same at any one time. So the potential difference across each bulb is always 1.2 V (12 V /10 bulbs).

00Picture1When 1 bulb is lit up from cold, with all the other 9 already hot, the initial potential difference across the cold bulb is much less than 1.2 V. This is because the other 9 hot filaments, with their larger resistances, has more than 1.2 V across each of them, as the potential divider principle dictates.

The pd across the 10th bulb will however eventually grow to become 1.2 V, when the bulb has heated up and its resistance becomes the same as all the other bulbs.

Compared to a bulb with 1.2 V across right from the beginning, a bulb with pd increasing slowly towards 1.2 V takes a longer time to light up.

002 How can a 2V EMF outsine a 3V EMF?

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While a battery provides the EMF (E) to push current through the external circuit, it also comes with its own internal resistance (r). The amount of EMF spent to push current through its own internal resistance is directly proportional to the current passing through the battery (Ir). So the amount of EMF left to push current through the external circuit, called the terminal potential difference (Vt), decreases whenever the battery is made to pump out larger current.

In this video, as we connect more and more bulbs (in parallel) across the battery, we draw larger and larger current from the battery. This must lead to a larger and larger potential difference across the internal resistance of the battery, which implies a smaller and smaller terminal pd, which dims the bulbs.

There is also a simple explanation why the 2V battery could “outshine” the 3V. The 2V battery may have a lower EMF, but by having a smaller internal resistance, it provided a higher terminal pd to the external circuit.

001 Why does the electron beam jump over the LDR (Light Dependent Resistor)?

 

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The circuit used in this demo is a simple potential divider consisting of an LDR (Light dependent resistor) and a fixed resistor. The potential difference across the fixed resistor is monitored on the CRO.

By facing the LDR towards the CRO screen, whenever the CRO trace arrives near the LDR, it illuminates the LDR and reduces its resistance. The potential difference across the fixed resistor thus increases. (This can be deduced by either (1) noting that the current in the circuit has increased or (2)the potential divider principle.) The CRO trace thus “jumps” over the LDR.

In a way, the CRO has ended up measuring its own brightness.