Electricity
One end of Nichrome wire of length 2L and cross-sectional area A is attached to an end of another Nichrome wire of length L and cross-sectional area 2A. If the free end of the longer wire is at an electric potential of 8.0 volts, and the free end of the shorter wire is at an electric potential of 1.0 volt, the potential at the junction of the two wires is most nearly equal to
A. 2.4 V
B. 3.3 V
C. 4.5 V
D. 5.7 V
E. 6.6 V
Cheers.
A. 2.4 V
B. 3.3 V
C. 4.5 V
D. 5.7 V
E. 6.6 V
Cheers.
Original post by Joinedup
Where did you get stuck?
it's a potential divider with a pd of 7 volts across it (8V-1V)
so if you can solve that you know the pd between the junction and the 1V line... which is 1V more than the pd between the junction and 0V
it's a potential divider with a pd of 7 volts across it (8V-1V)
so if you can solve that you know the pd between the junction and the 1V line... which is 1V more than the pd between the junction and 0V
The question just confused me. All i could do was draw the diagram and then was stuck.
Original post by 123chem
The question just confused me. All i could do was draw the diagram and then was stuck.
I might have got it now. Resistance for wire with length 2L and area A will be two times, and the other wire will have 0.5 times the resistance.
So, (8-V)/2 = I and (V-1)/0.5 = 1. Solve the simultaneous which gives 2.4. Is this right? Is there another method too?
Original post by 123chem
The question just confused me. All i could do was draw the diagram and then was stuck.
Well to get started you just need the ratio of the resistances of the long and short wires - which you can work out because they are made of the same material i.e. have same resistivity.
Original post by Joinedup
Well to get started you just need the ratio of the resistances of the long and short wires - which you can work out because they are made of the same material i.e. have same resistivity.
Done that, look at post above
Original post by 123chem
I might have got it now. Resistance for wire with length 2L and area A will be two times, and the other wire will have 0.5 times the resistance.
So, (8-V)/2 = I and (V-1)/0.5 = 1. Solve the simultaneous which gives 2.4. Is this right? Is there another method too?
So, (8-V)/2 = I and (V-1)/0.5 = 1. Solve the simultaneous which gives 2.4. Is this right? Is there another method too?
well the key is that being in series the current is the same in both.
the ratio of the resistances is 4:1
pd across the larger resistance is 4Vin/5
pd across the smaller resistance is 1Vin/5
and either subtract from 8V (or add to 1V as appropriate) to get your answer
---
or you could use the potential divider equation if you've memorised it
Vout = Vin R2/(R1+R2)
but important to remember that Vout is not relative to 0V... this question seems to have been designed to trick people who memorised the potential divider equation without really understanding it.
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