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Circuit problem

http://imgur.com/a/8smal
When the switch is open, the voltmeter reads 12.0V and when the switch is closed, it reads 8.0V
How to find the internal resistance r?

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

Just quoting in Danny Dorito so she can move the thread if needed :wizard:

Spoiler

Reply 2

Joinedup

Original post by Shinghei

http://imgur.com/a/8smal
When the switch is open, the voltmeter reads 12.0V and when the switch is closed, it reads 8.0V
How to find the internal resistance r?

Well the open circuit potential is 12V so the EMF (curly E) is 12 V
because no current is flowing in the internal resistance no voltage is present across the internal resistance and the voltmeter measures the full EMF

with the switch closed the EMF is the same 12.0V as before
but now a current is flowing in the internal resistance and in the external 4.0ohm resistance - the current in both must be the same because they are in series.

The PD across the internal resistance plus the PD across the external 4.0 ohm resistance must equal the EMF of 12.0V so the PD across the internal resistance must be 12.0 minus 8.0 Volts

The PD across the external resistance of 4.0 ohms is 8.0V so we can work out the current using I=V/R

The same current is flowing in the internal resistance and we need to find the value of r that gives us the correct PD across the internal resistance.

Reply 3

CodeX420

Original post by Shinghei

http://imgur.com/a/8smal
When the switch is open, the voltmeter reads 12.0V and when the switch is closed, it reads 8.0V
How to find the internal resistance r?

ok so I know its been 6 years but for future reference:

the open switch has 12v, but the close switch has 8 v - this means that the p.d "dropped" across the internal resistor is 4v (12v - 8v = 4v)

"V = IR"
so we calculate the current across the resistor (this will the current across all component since they're in series):
v/r = I --> 8/4 = 2A across the 4ohms resistor ( I use 8 becasue since we stated 4v is lost across internal resistor, the 4ohms resistor has p.d of 8v)

now finally to find the internal resistance we use the 2A current and the deduction that 4v was "dropped across the internal resistor:
v= 4
I = 2
therefore internal resistance (r) = 4/2 = 2ohms