Potential difference across one resistor with and without a known current?

Question

Okay imagine you have a voltage supply of 10V and one Resistor of 5 Ohms.

Now find the current : I=5/10 = 0.5A

So potential difference across that resistor : V=IR=10V .. which proves this statement I found in my lecture note :

If no internal resistance is present in voltage supply, the potential difference across the resistor is equal to supply voltage.

Now imagine the same circuit but total current is given as 0.1A . The potential difference is V=IR=0.1*5=0.5V, which basically means the statement above is incorrect.

However, my question is regarding the supplied current : what does it mean when the current is given or the need to find it ourselves ?? Is it possible to have the same voltage supply but with a different current supply ?

Let me know if you need more clarification

Answer 1

V=IR=10V .. which proves this statement I found in my lecture note

If no internal resistance is present in voltage supply, the potential difference across the resistor is equal to supply voltage.

You don't need to measure anything to prove this statement. It's a simple consequence of Kirchoff's voltage law. If you have a perfect 10 V voltage supply, no matter what you connect across it, the voltage across that element will be 10 V.

Now imagine the same circuit but total current is given as 0.1A . The potential difference is V=IR=0.1*5=0.5V, which basically means the statement above is incorrect.

I'll assume you know that your supply has an open-circuit voltage of 10 V, but you don't know the internal resistance.

If you measure 0.1 A, then you know the total resistance is 100 Ohms. This total resistance is made up of the supply's internal resistance and your external load (5 Ohms). Therefore you know the internal resistance is 95 Ohms.

Answer 2

Okay imagine you have a voltage supply of 10V and one Resistor of 5 Ohms.

Now find the current : I=5/10 = 0.5A

Wrong: if V = 10 V and R = 5 Ohms, then the current is given by Ohm's law:

$$ I = \frac{V}{R} = \frac{10}{5} = 2 \mathrm{\, A} $$

That, given that the voltage supply is ideal, which means that it doesn't have any internal resistance.

If no internal resistance is present in voltage supply, the potential difference across the resistor is equal to supply voltage.

This means that if the source is not ideal, it will show a series resistance which will add to the load causing a voltage divider that will lower the voltage on the load and dissipate some power.

It will look like this:

The voltage is given by:

$$ V_{load} = V_{supply} \cdot \frac{R_{load}}{R_{supply} + R_{load}} $$

Now, let's assume your book tells you that the current is now 0.1 A. The supply voltage will be still 10 V, and your load will still be 5 Ohms. But you now have a non-ideal supply, which will have the series resistance. Then your current is given by:

$$ I = \frac{V_{supply}}{R_{supply}+R_{load}} = 0.1 \mathrm{\, A} $$

and therefore

$$ R_{supply} = \frac{V_{supply}}{I} - R_{load} = \frac{10}{0.1} - 5 = 95 \mathrm{\, \Omega} $$