Help with Internal Resistance

In this video:

[If the embedded link doesn't work: https://www.youtube.com/watch?v=-3jZP5OQ-NQ]

I'm wondering why he does (V1 - V2) / V2 x by the Resistance of the Resistor...

I'm only familiar with the equation of:

V_external = EMF - (Amps X Internal Reistance)

Any help? I need to know this and be familiar with it for my next coursework :smile:

Thanks

Reply 1

uberteknik

Original post by Fangedbeast

In this video:

Thanks

Hello and welcome to TSR physics study help. :smile:

At what time in the video does this equation arise? I don't really want to look through the whole thing to find it!

Thanks.

Reply 2

lerjj

Original post by uberteknik

Hello and welcome to TSR physics study help. :smile:

At what time in the video does this equation arise? I don't really want to look through the whole thing to find it!

Thanks.

It's about 3 minutes in, V1 is voltage across the battery with no load attached, V2 has a resistance attached (they used a 100Ohm resistor, that shouldn't make a difference).

I'm having trouble seeing why that equation's correct as well. We measured internal resistance with a different method when we did it in class.

Reply 3

uberteknik

Original post by Fangedbeast

.........

I'm wondering why he does (V1 - V2) / V2 x by the Resistance of the Resistor...

I'm only familiar with the equation of:

V_external = EMF - (Amps X Internal Reistance)

Any help? I need to know this and be familiar with it for my next coursework :smile:

Thanks

Original post by lerjj

V_1

is the open circuit battery voltage and since the ohm meter has an extremely high input resistance (>>1x10⁶ ohms), virtually no voltage (<<1uV) will be dropped across the battery internal resistance when the meter takes an open circuit voltage measurement. The open circuit voltage can therefore be assumed equal to the battery E.M.F. to 6 d.p or more.

V_2

is the p.d. developed across the 100 ohm load.

From Kirchoff's voltage rule, the p.d. developed across the battery internal resistance (100 ohms load attached) is then

V_{int} = V_{emf} - V_{load}

i.e.

V_{int} = V_1 - V_2

.

Because the current in the load resistance must also flow through the battery internal resistance (series current path), we can say:

I_{int} = I_{load}

\frac{V_{int}}{R_{int}} = \frac{V_2}{R_{load}}

Substituting for

V_{int}

\frac{V_1 - V_2}{R_{int}} = \frac{V_2}{R_{load}}

then make

R_{int}

the subject.

\frac{V_1 - V_2}{V_2} \ \mathrm {x \ }R_{load} = R_{int}

QED.

(edited 9 years ago)

Reply 4

lerjj

Original post by uberteknik

V_1

How the hell did I not realise this? Everything else makes perfect sense now.

Reply 5

uberteknik

Original post by lerjj

How the hell did I not realise this? Everything else makes perfect sense now.

The thing with much of electronics is knowing how a measurement instrument works together with it's benefits and most importantly, it's limitations.

Students tend to pick up an instrument and assume the measurement is accurate to however many decimal places there are in the display, without understanding the implications of the accompanying equipment specifications.