Finding voltage drop in simple circuit.

[billybob]

Hello there, I have been trying to find a solution to a seemingly very simple problem. So with this circuit a battery and a resistor imagine I know the resistor value maybe 200 ohms or something and I know the voltage of the battery maybe 10 volts. If I attached a multimeter from one side of the resistor to the other what voltage would it read? I tried finding the current then multiplying it by the resistance, but I just get the input voltage (which is expected). I know how to solve this problem if I know what current and voltage I NEED after the resistor, but not this.

Please help, I am a beginner.
-Ben

 

[fourtytwo]

You have already stated the battery voltage is 10 volts and it clearly connected directly to the resistor so the answer is 10 volts across the resistor.

 

[billybob]

You have already stated the battery voltage is 10 volts and it clearly connected directly to the resistor so the answer is 10 volts across the resistor.

but if i had a load, I thought that the current drop due to a resistor is proportional to the voltage drop. Am I wrong?

 

[billybob]

also if I had an LED or something, 10 volts seems like way too much.

 

[MacIntoshCZ]

but if i had a load, I thought that the current drop due to a resistor is proportional to the voltage drop. Am I wrong?

If you connect it directly to power supply it will blow.
You need calacute resistor.
(VCC-Vforward-Rwires*Idiode)/Idiode)
No one care about wires Rwires. Rwires aprox 0 ohms
Profesional picture attached:

Also i remember your post, when you said you have good understanding in electronics parts.

Kirchhoff's circuit laws - Wikipedia

en.wikipedia.org

Ohm's law - Wikipedia

en.wikipedia.org

 

[audioguru]

but if i had a load, I thought that the current drop due to a resistor is proportional to the voltage drop. Am I wrong?

You are talking about the voltage dropping on an old overloaded battery.

 

[billybob]

If you connect it directly to power supply it will blow.
You need calacute resistor.
(VCC-Vforward-Rwires*Idiode)/Idiode)
No one care about wires Rwires. Rwires aprox 0 ohms
Profesional picture attached:View attachment 129811
Also i remember your post, when you said you have good understanding in electronics parts.

Kirchhoff's circuit laws - Wikipedia

en.wikipedia.org

Ohm's law - Wikipedia

en.wikipedia.org

I'm still not quite understanding how to output voltage can be found. I use the formula inV-neededV/current needed will result in the resistor required, but if the resistor is already known how is the resulting voltage found using ohms law. I can find the voltage drop of each individual resistor if there are two or more in series or parallel, but why does one resistor not apply?

Also yes I did say that. Kinda putting me in my place. Also that comment is subjective and I really don't know what a good understanding is when there is always more to learn.
Thank you for your time.
-Ben

 

[MacIntoshCZ]

Ok sorry my first comment coul be dissaponting. All diodes has forvard voltage proportional to passing current. Higher current means higher forward voltage. You need to take a look to datasheet. For emiting diodes i never do this. I just select forward voltage in range of 2-3V and current aprox 5mA. If you want exactly 5mA than you would need something like current mirror as current source. Also forward voltage on leds vary by colours

 

[billybob]

Ok sorry my first comment coul be dissaponting. All diodes has forvard voltage proportional to passing current. Higher current means higher forward voltage. You need to take a look to datasheet. For emiting diodes i never do this. I just select forward voltage in range of 2-3V and current aprox 5mA. If you want exactly 5mA than you would need something like current mirror as current source. Also forward voltage on leds vary by colours

I was just giving an LED as an example, what if it was an open circuit that was completed by a multimeter on the volts setting? What would it read?

 

[MacIntoshCZ]

I was just giving an LED as an example, what if it was an open circuit that was completed by a multimeter on the volts setting? What would it read?

You mean voltmeter in series with load? Then it depend on impedance of circuitry. It will act as resistor divider. Multimeters has input impedances in range of 10Megaohms? Maybe more.I dont remember exactly. In milivolts range impedance is the highest. Cause it can go directly to op amp. For higher voltages there is resistor divider inside multimeter
For low circuit impedance it Will read vcc

 

[billybob]

Here is a link talking about how to find the voltage drop, but it's only with 2 resistors. Is it not possible to find the voltage drop with just one resistor???

 

[Visitor]

Let's drop back to Ohm's Law: V = i/r, which can be rearranged as i = V/r or r = V/i, where V is volts, i is current in amps and r = Ohms.

In the circuit below, we have a battery, (which we will assume has a constant voltage regardless of load and no internal resistance), a resistor representing the circuit load and another resistor representing the resistance of the wiring and connections. The current, shown by the red line, is constant throughout the circuit, since everything is in series.

Let's say the battery provides 12 volts, the load has a resistance of 10 Ohms, and assume the wiring has 0 ohms resistance,

Use Ohm's law to find the current:

i = V/r = 12/10 = 1.2 amps.

If the cable resistance truly is zero Ohms, the voltage across the Load will be the same as the (ideal) battery voltage.

But let's say the voltage across the load only measures 11 volts. That means 1 volt is being dropped across the cable resistance, which is actually spread out rather than being a single lump like I've shown here. Back to Ohm's law. We can calculate the resistance in the cable.

We already know 1 volt is dropped across the cable resistance. But what's the current? The load is 10 Ohms we know, and there's 11 volts across it.

i = V/r = 11/10 = 1.1 amps.

The same current is flowing through the cable, so we know the voltage drop caused by the cable resistance, and the current, so

r = V/i = 1/1.1 = 0.91 Ohms = cable resistance.

Let's doublecheck our calulations.

Rload = 10 Ohms
Rcable = 0.91 Ohms

Rtotal = Rload + Rcable = 10 + 0.91 = 10.91 Ohms

i = V/r = 12/10.91 = 1.1 amps.

 

[MacIntoshCZ]

Here is a link talking about how to find the voltage drop, but it's only with 2 resistors. Is it not possible to find the voltage drop with just one resistor???

If you want to connect resistor directly to power supply than no. No voltage drop.
One of those two resistor can be considered as load. With variable resistence current vary and then voltage drop at one resistor vary too.

 

[Visitor]

If you have only a power supply and a load, the voltage across the load will equal the power supply voltage. But if we dig a little deeper, the cable has resistance (as I showed above) and a power supply will have internal resistance. This internal resistance is usually very small and usually insignificant. The lower this internal resistance, the more "stiff" a power supply is called,

 

[billybob]

Let's drop back to Ohm's Law: V = i/r, which can be rearranged as i = V/r or r = V/i, where V is volts, i is current in amps and r = Ohms.

In the circuit below, we have a battery, (which we will assume has a constant voltage regardless of load and no internal resistance), a resistor representing the circuit load and another resistor representing the resistance of the wiring and connections. The current, shown by the red line, is constant throughout the circuit, since everything is in series.

Let's say the battery provides 12 volts, the load has a resistance of 10 Ohms, and assume the wiring has 0 ohms resistance,

Use Ohm's law to find the current:

i = V/r = 12/10 = 1.2 amps.

View attachment 129812

If the cable resistance truly is zero Ohms, the voltage across the Load will be the same as the (ideal) battery voltage.

But let's say the voltage across the load only measures 11 volts. That means 1 volt is being dropped across the cable resistance, which is actually spread out rather than being a single lump like I've shown here. Back to Ohm's law. We can calculate the resistance in the cable.

We already know 1 volt is dropped across the cable resistance. But what's the current? The load is 10 Ohms we know, and there's 11 volts across it.

i = V/r = 11/10 = 1.1 amps.

The same current is flowing through the cable, so we know the voltage drop caused by the cable resistance, and the current, so

r = V/i = 1/1.1 = 0.91 Ohms = cable resistance.

Let's doublecheck our calulations.

Rload = 10 Ohms
Rcable = 0.91 Ohms

Rtotal = Rload + Rcable = 10 + 0.91 = 10.91 Ohms

i = V/r = 12/10.91 = 1.1 amps.

You explained that very well. Thank you. I guess I thought voltage was way more effected by resistance than it really is. So I know your banging your heads against the table right now, but just help me clarify this... If I had say, 100 volts as the source going through wire (imagining no voltage drop) going across a 500 ohm resistor like the previous picture but this time there is a load adding which is a lamp or something that draws a little more current, would the voltage read STILL remain 100 volts? Even when resistance is increased and increased? My understanding of Kirchhoff's law is the voltage drops add to equal the input voltage. Is current and voltage not effected by the other??? Please give me a new perspective of my flawed view of this. Or maybe point me where I can learn.

Thank you, Ben

 

[Visitor]

If the lamp is added in parallel to the load resistor, the voltage across both would be the same. But the combined resistance would be lower (look up adding resistors in parallel), so the current would be higher from the power supply.

If the lamp was in series with the load resistor, the effective resistance is the sum of the two resistances, and the current through the circuit would drop accordingly.

 

[audioguru]

100V then 120V somewhere?
500 ohm resistor connected in series or in parallel with a light bulb?
Increase and increase a resistance somewhere?

A schematic explains everything but you did not post a schematic so we don't know what you are talking about.

 

[billybob]

If the lamp is added in parallel to the load resistor, the voltage across both would be the same. But the combined resistance would be lower (look up adding resistors in parallel), so the current would be higher from the power supply.

If the lamp was in series with the load resistor, the effective resistance is the sum of the two resistances, and the current through the circuit would drop accordingly.

OOOOOOOOHHHHH. I just had a light bulb moment!!! there is always two resistors!!!

 

[billybob]

In series with the lamp the lamp might be 2 ohms and in series with the 500 ohm resistor I can then use the simple ohms law equation to find the voltage drop after each resistor!!

 

[billybob]

100V then 120V somewhere?
500 ohm resistor connected in series or in parallel with a light bulb?
Increase and increase a resistance somewhere?

A schematic explains everything but you did not post a schematic so we don't know what you are talking about.

Oh yeah your right. Fixed it.