Calculate voltage drop by using kirchoff's law

[Arno@praxispc]

Homework Statement

The circuit has a battery Voltage of 30V, a resistor of R1(8ohms), R2 (2Ohms) and R3 (4ohms) in series. Calculate the voltage drop over R1

Homework Equations

Please give me the proper equation of this? The one I've got says you take 30-8I-2I-4I=0
Wich gives you current of 2.142 and not the voltage.

The Attempt at a Solution

Now what i tried doing is this.
30V-8-2-4=0
30V=14
V=14/30
V=0.466
but I'm sure its wrong.

 

[vk6kro]

You have the correct current, although you could just add the resistor values up using mental arithmetic and work out the current using Ohm's Law.

This current flows through an 8 ohm resistor, so what is the voltage across the resistor?

 

[Arno@praxispc]

ok so what you are saying is that if i use ohms law V=I*R
I will have to use the 8 ohm resistor times by the current of 2.142= 17.136V
Is this the voltage after it went through the resistor or is this the amount of voltage drop?

 

[vk6kro]

The voltage across a resistor is equal to the current through it times the resistance of that resistor.

So, the 8 ohm resistor times the current of 2.142 Amps = 17.136 Volts.

If you do this for all the resistors, you will see that they add up to the supply voltage of 30 volts.

You should also see that the bigger resistors drop more of the the available voltage than the smaller ones, exactly in proportion to their resistance.

 

[DeeNos]

To calculate the voltage drop over a specific resistor in a series circuit, you can use Kirchoff's Voltage Law (KVL). This law states that the sum of the voltage drops around a closed loop in a circuit is equal to the sum of the voltage sources in that loop. In this case, we have one voltage source (the battery) and three resistors in series. To find the voltage drop over R1, we can use the equation V1 = Vbattery - V2 - V3, where V1 is the voltage drop over R1 and V2 and V3 are the voltage drops over R2 and R3 respectively.

Plugging in the values given in the problem, we get V1 = 30V - (8ohms)(2.142A) - (2ohms)(2.142A) - (4ohms)(2.142A) = 30V - 17.136V = 12.864V. Therefore, the voltage drop over R1 is 12.864V.

Your initial attempt at solving the problem was on the right track, but you made a few mistakes. First, when using KVL, it is important to use the correct signs for the voltage drops. Since the current is flowing in the same direction through each resistor, the voltage drop over each one will be negative (since the current is flowing against the direction of the voltage). So, the equation should have been 30V - (-8I) - (-2I) - (-4I) = 0. This would give you the correct answer of 12.864V.

Additionally, in your second attempt, you divided the total voltage by the total resistance, which is not the correct approach. This would give you the total current in the circuit, but not the voltage drop over a specific resistor.

I hope this helps clarify the process for solving this type of problem. Remember to pay attention to the signs and units when using KVL.