Hedgeman49
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#1
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Incredibly annoying question which I just don't know how to approach. (may be hard to convey without the diagram, but I'll try. I have written solutions to the questions I have already completed.)

a) A cell of emf \epsilon and internal resistance r is connected in series to a resistor of resistance R as shown. A current I flows in the circuit.
(Shows standard series circuit with said components.)

i) State an expression which gives \epsilon in terms of I, r, and R.
Answer: \epsilon = I(R+r)

ii) Hence show how V_r, the p.d. across the resistor, is related to \epsilon, I and r.

Answer: V_r = \epsilon - Ir

b) A lamp, rated at 30W, is connected to a 120V supply.
i) Calculate the current in the lamp.
Answer: I=\frac{P}{V}=\frac{30}{120}=0.25A

ii) If the resistor in part (a) is replaced by the lamp described in part (b), determine how many cells, each of emf 1.5V and internal resistance 1.2\Omega, would have to be connected in series so that the lamp would operate and its proper power.
Attempt at answer:
R of lamp =\frac{V}{I}=\frac{120}{0.25}=480\Omega
I=\frac{\epsilon}{R+r}
I=\frac{1.5}{480+1.2}
I=3.12x10^{-3}A
V=\frac{P}{I}=\frac{30}{3.12x10^{-3}}=9624
\RightarrowNumber of cells = \frac{9624}{1.5}=6416, which is quite clearly a ridiculous figure, so there must be a hole in my understanding somewhere... would anybody care to fill it? This question has been getting at me all day long.
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(Original post by Hedgeman49)
Incredibly annoying question which I just don't know how to approach. (may be hard to convey without the diagram, but I'll try. I have written solutions to the questions I have already completed.)

a) A cell of emf Image and internal resistance r is connected in series to a resistor of resistance R as shown. A current I flows in the circuit.
(Shows standard series circuit with said components.)

i) State an expression which gives Image in terms of I, r, and R.
Answer: Image

ii) Hence show how Image, the p.d. across the resistor, is related to Image, I and r.

Answer: Image

b) A lamp, rated at 30W, is connected to a 120V supply.
i) Calculate the current in the lamp.
Answer: Image

ii) If the resistor in part (a) is replaced by the lamp described in part (b), determine how many cells, each of emf 1.5V and internal resistance 1.2Image, would have to be connected in series so that the lamp would operate and its proper power.
Attempt at answer:
R of lamp =Image
Image
Image
Image
Image
ImageNumber of cells = Image, which is quite clearly a ridiculous figure, so there must be a hole in my understanding somewhere... would anybody care to fill it? This question has been getting at me all day long.
You need I think:

P=30W = I^2R = \left( \frac{1.5n}{480+1.2n}\right)^2\times 480

and solve for n where n is the number of cells (I get n=100)
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Glutamic Acid
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I'm not too confident on electricity, but ...
P = IV
30 = I(E - Ir)k, where k is the number of cells.
30 = 0.25(1.5 - 0.25 x 1.2)k, and solve...

However, there is a large chance I am wrong.
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(Original post by Glutamic Acid)
I'm not too confident on electricity, but ...
P = IV
30 = I(E - Ir)k, where k is the number of cells.
30 = 0.25(1.5 - 0.25 x 1.2)k, and solve...

However, there is a large chance I am wrong.
no you're right, it gives the same answer. I just did it in such a way as to follow on from the working already there.
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Hedgeman49
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Cool, thanks a lot guys. Feel kinda ashamed I didn't spot that one! Bit of a stupid question if you ask me... who would put 100 cells in a circuit?
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Theorist
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(Original post by Hedgeman49)
who would put 100 cells in a circuit?
An AQA exam writer. :p:
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true, a big number or not, it's always encouraging when you come out with nice round numbers :p:
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Theorist
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Indeed.
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tara dawood
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thanks
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tara dawood
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thanks - tara
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tara dawood
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thanks
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