circuits in which there are 2 sources of e.m.f
Why does the resistance increase when there are fewer components switched on?
This is a diagram from my book:
Then they say that 'if you switch off some of the car's equipment, there is an increase in the external circuit's resistance from 1.2 ohm to 5 ohm'.
But I don't understand why this happens. I understand the maths but not the concept, I mean I know that because the batteries aren't providing any current and as it takes current instead only 2.4 amps reaches the components so the resistance is now: R = V/I, and as the voltage remains the same then: R=12 /2.4 = 5. But like I said I don't understand why this is the case.
This is a diagram from my book:
Then they say that 'if you switch off some of the car's equipment, there is an increase in the external circuit's resistance from 1.2 ohm to 5 ohm'.
But I don't understand why this happens. I understand the maths but not the concept, I mean I know that because the batteries aren't providing any current and as it takes current instead only 2.4 amps reaches the components so the resistance is now: R = V/I, and as the voltage remains the same then: R=12 /2.4 = 5. But like I said I don't understand why this is the case.
(edited 10 years ago)
Original post by tammie94
Why does the resistance increase when there are fewer components switched on?
This is a diagram from my book:
Then they say that 'if you switch off some of the car's equipment, there is an increase in the external circuit's resistance from 1.2 ohm to 5 ohm'.
But I don't understand why this happens. I understand the maths but not the concept, I mean I know that because the batteries aren't providing any current and as it takes current instead only 2.4 amps reaches the components so the resistance is now: R = V/I, and as the voltage remains the same then: R=12 /2.4 = 5. But like I said I don't understand why this is the case.
This is a diagram from my book:
Then they say that 'if you switch off some of the car's equipment, there is an increase in the external circuit's resistance from 1.2 ohm to 5 ohm'.
But I don't understand why this happens. I understand the maths but not the concept, I mean I know that because the batteries aren't providing any current and as it takes current instead only 2.4 amps reaches the components so the resistance is now: R = V/I, and as the voltage remains the same then: R=12 /2.4 = 5. But like I said I don't understand why this is the case.
Its because all the components connected to the battery are in parallel and will all draw their own current. The current is taken from the same source so they all add up. Iload = Iradio + Iaircon + Iheadlamp + Ietc.....
The 'equivalent' resistance of all those components in parallel is calculated from Rload = Vbatt/Iload
Where Rload = Rradio llel Raircon llel Rheadlamp llel Retc.....
When some of the components are switched off, less current is taken from the battery. And from the same equation above the equivalent load resistance must therefore increase.
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