# AQA AS Physics Unit 3 - EMF

WatchPage 1 of 1

Go to first unread

Skip to page:

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 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 in terms of I, r, and R.

Answer:

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

Answer:

b) A lamp, rated at 30W, is connected to a 120V supply.

i) Calculate the current in the lamp.

Answer:

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, would have to be connected in series so that the lamp would operate and its proper power.

Attempt at answer:

R of lamp =

Number of cells = , 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.

a) A cell of emf 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 in terms of I, r, and R.

Answer:

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

Answer:

b) A lamp, rated at 30W, is connected to a 120V supply.

i) Calculate the current in the lamp.

Answer:

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, would have to be connected in series so that the lamp would operate and its proper power.

Attempt at answer:

R of lamp =

Number of cells = , 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.

0

reply

Report

#2

(Original post by

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 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 in terms of I, r, and R.

Answer:

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

Answer:

b) A lamp, rated at 30W, is connected to a 120V supply.

i) Calculate the current in the lamp.

Answer:

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, would have to be connected in series so that the lamp would operate and its proper power.

Attempt at answer:

R of lamp =

Number of cells = , 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.

**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 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 in terms of I, r, and R.

Answer:

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

Answer:

b) A lamp, rated at 30W, is connected to a 120V supply.

i) Calculate the current in the lamp.

Answer:

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, would have to be connected in series so that the lamp would operate and its proper power.

Attempt at answer:

R of lamp =

Number of cells = , 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.

and solve for n where n is the number of cells (I get n=100)

0

reply

Report

#3

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.

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.

0

reply

Report

#4

(Original post by

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.

**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.

0

reply

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?

0

reply

Report

#6

(Original post by

who would put 100 cells in a circuit?

**Hedgeman49**)who would put 100 cells in a circuit?

1

reply

Report

#7

true, a big number or not, it's always encouraging when you come out with nice round numbers

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top