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What will be the $V-I$ graph of a Bulb?

APPROACH : I think as the potential difference across a filament bulb increases, the current increases and the energy dissipated, as heat, increases, resulting in a higher temperature. As the temperature increases, resistance of the filament increases. The collision between the free electrons and the lattice ions increases due to more rigorous vibration of the lattice ions .

So my conclusion is:

enter image description here

$\Rightarrow$ But I'm not able to provide a valid reason to this , help me with this .

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Your reasoning appears to be valid, and this is typical non-Ohmic behavior for an incandescent light. Non-Ohmic meaning a non-linear relationship between voltage and current.

Initially, as the potential difference across the filament is increased, so to does the current inside it and energy is dissipated as heat. This results in the filament operating at a higher temperature. But as you have also pointed out, the temperature increases, which results in the resistance of the filament also increasing.

If we have a look at the current-voltage current graph for such a process, we get something like this:

enter image description here

Although the graph you drew was voltage versus current, it is similar to this which shows current versus voltage.

Such non-Ohmic behavior results because of the heat generated by the filament in the lamp. For a lamp powered by a normal battery, for example a torch, such a power sources provides (nearly) constant voltage.

When first turned on, the incandescent lamp starts with a relatively low resistance, like a few Ohms. Because of this, there will be a large in-flow of current. This means that the filament will become hot very quickly. With this increase in temperature, the resistance rises, ultimately resulting in a reduced current as the lamp settles to its normal operation. As you have described, the lattice vibrations lead to an enhanced scattering of electrons which reduces the current flow.

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I posted this question six years ago which contains experimental graphs of V vs I. In the question I discuss the fact that

  1. Filament bulbs are non-linear in their characteristic VI curve.
  2. The resistance is initially low and increases with temperature.
  3. A low frequency AC signal will show a hysteresis.

You may be interested to see a "backbend" (Limbo?) in the current as the voltage magnitude is increased. As the voltage magnitude decreases, the curve follows the smoother path back to a ($0$ V, $0$ A) point. This curve is taken at an AC frequency of 2 Hz, IIRC. enter image description here Also, though not shown in the question, at higher frequencies (>50 Hz) the VI curve will become linear because the temperature of the filament reaches an equilibrium value.

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