# Voltage across a 2.2k resistor [closed]

I've just begun learning about electronics, and have been performing a few experiments on measuring voltage,current with a multi-meter.

If I have a single resistor having a resistance of 2.2k connected to a 9v battery(without any other load within my circuit), when I apply the probes of the multi meter in series with the circuit, shouldn't the multi-meter show a voltage drop? As currently it shows 9V.

I say so, as If I measure the current, it results in a drop of 5mA from 0.5A. Could anyone elaborate on why such a behavior?

• A register is a digital logic element that stores digital values. A resistor is a circuit element that allows current to flow in proportion to the voltage across it. – The Photon Jul 21 '16 at 5:10
• lol adding the register ... do you mean you read the current with the resistor and it was 5mA. And you read the current without the resistor and it read 500mA (.5A)? – jbord39 Jul 21 '16 at 5:14
• When you connect an ammeter across a battery, you're basically measuring the short circuit current of the battery, limited by the battery's internal resistance. It appears that your battery's internal resistance is about 18 ohms. This becomes mostly irrelevant when you put it in series with 2200 ohms. We don't call this a "current drop". We just call it a different current flowing in a different circuit. – The Photon Jul 21 '16 at 5:14
• You came here, ostensibly, to learn. But after being told twice that a register and a resistor are different things, you continue to call a resistor a register. Now I think we're just being trolled. – The Photon Jul 21 '16 at 5:28
• I'm voting to close this question as off-topic because the asker is deliberately posting nonsense. – The Photon Jul 21 '16 at 5:31

Ohms law describes the voltage, current and resistance relationship.

$$V = IR$$

One way to read this would be "The voltage across a resistor is equal to the current through the resistor multiplied with its resistance".

So in the case of a 2.2k resistor hooked up to a 9V battery, there is a voltage drop of 9V and that's what your multi meter is telling you.

To illustrate how all this works lets work out the current in this case. In terms of ohms law we know the resistance of the resistor and voltage V across the resistor to be 9V since it's connected to the 9V battery. So according to Ohms law we have:

$$9 = 2200I$$ $$I=\frac{9}{2200}$$ $$I = 4.1mA$$

As another example, lets say we have a 9V battery connected to a 1k resistor which is then connected in series to a led. Lets say the forward voltage (voltage drop across a diode when conducting) of the led is 3V. What is the voltage across the resistor and the current through it?

Well we know that the battery is providing 9V and the led drops 3V so this leaves 6V across the resistor. So the current through the resistor is $$I = \frac{V}{R}$$ $$I = \frac{6}{1000}$$ $$I = 6mA$$

In general, out of the three parameters V, I and R you can only control two of them at once. The third parameter is a result of the other two.

For example if you set the voltage and resistance, the current will be derived. If you set the voltage and current, the resistance will be derived. If you set the current and resistance, the voltage will be derived.

Firstly, it seems that you measured the curent by shorting the battery with your measuring device, don't do that. Current measuring is done in series with the load. There is no voltage drop because you effectively create a voltage divider with the internal resistance of the multimeter on the voltage scale, which is in the Megaohm range. So actually there is a voltage drop, but is so small that you won't be able to see it.

• But how is it that I can measure the drop when I added a LED along with a register. How does this differ? – Akash Jul 21 '16 at 5:38
• Because the led has a voltage drop. – user1657930 Jul 22 '16 at 6:27

"When I apply the probes of the multi meter in series with the circuit, shouldn't the multi-meter show a voltage drop? As currently it shows 9V."

The 9V is your voltage drop, using KVL we know the voltage supplied is the same as the sum of all the voltage drops in the circuit. If you've only got the one component (The 2.2k resistor) then all the voltage will be dropped across that one component, hence why it is the same voltage as your supply voltage.

Try the following: Add another 2.2k resistor in series and then measure the voltage across one. You should see half the supply voltage 