Question

A resistance "a" of 200 ohms is connected in series to a battery of 110 volts e.m.f and 20 ohms internal resistance. Find the terminal potential difference of a. If the potential difference is measured by the use of a voltmeter of 8000 ohms resistance, find the error in the voltmeter's reading.

Answer 1

Treat the real battery as if it was an ideal battery in series with a 20 ohm resistor.

Answer 2

I've calculated the terminal potential difference of a, using V/R+r. It equals 0.5 amperes. Now comes the second part, that part I can't figure out.

Answer 3

0.5 amperes is the current flowing in the circuit in the first part, but what figure do you get for the terminal pd ?

Answer 4

Yeah sorry my bad. The current equals 0.5A, so the terminal pd equals 100, through V = Vb-Ir.

Answer 5

ok, right
so now, if you connect up the voltmeter, you're adding 8000 ohms in parallel with the 200 and that will affect the current flowing, which in turn will give you a different terminal pd

Answer 6

You mean in parallel with 220 right? Since R total is 200 + 20 = 220.

How do I calculate the terminal pd then? Do I use the same current of 0.5A?

Answer 7

no, the meter is in parallel with the 200 - think about where you would actually connect the voltmeter in the real circuit

Answer 8

Alright so the new resistance is 8200ohms? How do I calculate the terminal pd now?

Answer 9

it would be 8200 if the meter was connected in series, but it's connected in parallel, different

Answer 10

basically, in the first circuit the real battery had 200 ohms across it
now with the voltmeter connected as well, the real battery 'sees' 200 in parallel with 8000

Answer 11

that will give an effective resistance that is slightly less than 200, so the current goes up
first you need to calculate the effective resistance of the 200 and 8000 in parallel, then use that to figure out the new current in the circuit

Answer 12

Oh I'm sorry I'm not thinking enough because I'm in a hurry. So 8000 and 200 in parallel will equal 1/R = 1/8000 + 1/200 which will equal 1/195.121

So the new R is 195.121

Answer 13

yes

Answer 14

so now you see the current is slightly different and therefore the voltage dropped across the internal resistance changes

Answer 15

now that you have the 195 ohms, you just need to recalculate the voltage across the battery terminals like you did in the first part

Answer 16

I have a current of 0.5A, so now all I need to do is V = IR = 0.5 x 195 = 97.6v approx

So the error is 2.4 approx?

Answer 17

why do you have a current of 0.5 amps ? the new resistance changes the current

Answer 18

you have to recalculate the current and remember the 20 ohms internal resistance of the battery

Answer 19

I need to calculate the potential difference to know the error in reading..

Answer 20

which potential difference ?

Answer 21

I need to get the reading of the voltmeter, re-read the problem.

Answer 22

if you just type 'the potential difference', i don't know which potential difference you are referring to, so i asked for clarification to avoid confusion

Answer 23

Alright, sorry for the confusion.

Answer 24

so where are you up to now, you have the 195 and you need to figure out the current now flowing

Answer 25

or you could just treat the 20 and 195 like a potential divider, whichever you find easier
yes, we are aiming at finding the pd across the battery terminals in the new situation then you can see the change from the 100 volts in part 1

Answer 26

So, I need to calculate the new current intensity using the new resistance.

I = Vb/R + r = 110/8200 = 0.01341 A

Then calculate the new terminal pd (V)
V = Vb - Ir = 110 - (0.01341 x 20) = 109.7v

Is that correct?

Answer 27

no, you're not getting the parallel resistance 8200 would be the SERIES resistance, you have 200 and 8000 in PARALLEL
and you still have to add in the 20 ohms internal resistance after the parallel calculation

Answer 28

I forgot about it again, sorry. So:

I = Vb / R + r = 110 / (195.121 + 20) = 0.51134 A
V = Vb - Ir = 110-(0.51134 x 20) = 99.7731v
So the error is 0.22 is that correct this time?

Answer 29

yes, you got it, i think it was closer to 0.23 volts error to 2dp's

see if you can see how an ideal voltmeter would have infinite internal resistance from this example