Mastering physics question!!!!

topsquark

Forum Staff
Apr 2008
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656
On the dance floor, baby!
Using Ohm's Law we see that \(\displaystyle R = \dfrac{ \Delta V}{I}\), which is 1/slope. You can actually calculate what the R's are but we can eyeball it. For example at \(\displaystyle \Delta V = 3 ~ V\) we see that the slope for the resistor is larger than the slope for the lightbulb. Thus the resistance in the resistor is less than the resistance in the lightbulb, which is option B. So you got that one right.

What can you do for the other possible answers?

-Dan
 
Apr 2015
1,219
349
Somerset, England
This is a very good way to present your question.

I'd be inclined to assume that the two curves come together at 0 volts and 3 volts.

This would be consistent with the three sections of the questions.

Topsquark is correct to say use Ohms law, but I would do it differently, since comparing slopes can be misleading.

So for any V between 0 and 3 volts (ie 0 > V > 3) The current in the bulb is greater than the current in the resistor or Ib > Ir.

So for any of these values Δv is the same so if we take a vertical line upwards on the plot from any ΔV value we meet the resistor curve before the bulb curve.

But ΔV is the same for both bulb and resistor at this value so ΔV/Ir > ΔV/Ib ; where Ir and Ib are the resistor and bulb currents respectively.

The resistor resistance will alway be greater than the bulb resistance in this Δvoltage range.

For example
looking at the curve at ΔV = 2 volts, Ib = 125mA and Ir = 100mA

So R(bulb) = 2/0.125 = 16 ohms
and R(resistor) = 2/0.1 = 20 ohms

At ΔV = 3 volts I make the resistances equal.

That's two parts, what do you think about the third part?
 
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