[sabre2th1]

The answer is C, why isn't it 9/20 = 0.45 ?

[Stonebridge]

(Original post by sabre2th1)


The answer is C, why isn't it 9/20 = 0.45 ?

Because the resistance in the circuit is not just 20 ohms, there is also the bulb.

The way to do this is to look at the graph and decide which of the currents along the bottom have the two voltages (bulb + resistor) adding up to 9V.
The sum of the pds across the two must be 9V as they are in series.

[AtomSmasher]

Is this GCSE or something else?

[sabre2th1]

(Original post by Stonebridge)
Because the resistance in the circuit is not just 20 ohms, there is also the bulb.

The way to do this is to look at the graph and decide which of the currents along the bottom have the two voltages (bulb + resistor) adding up to 9V.
The sum of the pds across the two must be 9V as they are in series.

But if you look at the graph, there are many different combinations that add up to 9V. (eg 6+3, 4+5, 2+7) ?

[sabre2th1]

(Original post by AtomSmasher)
Is this GCSE or something else?

AS physics

[Pangol]

(Original post by sabre2th1)
But if you look at the graph, there are many different combinations that add up to 9V. (eg 6+3, 4+5, 2+7) ?

What Stonebridge is saying is that you have to find a value for the current where the potential difference across the resistor and the potential difference across the filament lamp add up to 9 V - for the same value of the current.

You are right to say that, for example, 2 + 7 = 9, but when the potential difference across the resistor is 2 V (at a current of 0.1 A), the potential difference across the lamp is only 0.8 V, and 2 + 0.8 is not 9. (Or, alternatively, when the potential difference across the lamp is 2 V, the potential difference across the resistor is about 4.4 V, and again, 2 + 4.4 is not 9.)

There is only one place where, for the same current, the potential differences across the two components add up to 9 V.