Voltage and Resistance, please help?

Question

Please help with the following exercise:

Given 2 resistors in series R1=10 ohms, R2= 30 ohms, determine the current flowing through the resistors and the voltage across each resistor, if the potential difference across the wire is V=12V.

This is how I solved it:

The total resistance Req= R1+R2= 40 ohms

R = V / i
i = V/R = 0.3 A

V(R1) = -R1*i = -10*0.3= -3V
V(R2) = -R2*i = -30*0.3= -9V

Is it correct? Can the voltage be negative? Please help.

Thank you.

Answer 1

I am almost certain that you did it almost all correctly. However, when I learned it I did not learn it with the negative sign[Yours appears to be V=(I)(-R)].
V=IR is the formula, as I'm sure you know. After finding the current, as you did correctly, you would plug it into the formula using the resistance of each resistor.
V1=(.3)(10)=3
V2=(.3)(30)=9
If your formula has a negative, it would be- I suppose-only there to remind you that it is a voltage DROP. You would lose that much V through each resistor.
In the end, the voltage gained should equal the voltage dropped, which it does. 9+3=12

(This is slightly different in a parallel arrangement because the voltage of resistors in a series is the same but the current changes)

I hope this was helpful.

Answer 2

The voltage drops across the entire wire, so the fraction that drops across each resistor must be
V(R1) = V* R1/(R1+R2) = 12 * 10(40) = 3 V
V(R2) = 12 V - 3 V = 9 V OR V*R2/(R1+R2) = 12*30/(40) = 9V

The current is as you state
V = I*R
I = V/R = 12/(40) = 0.3 A

Your method would be fine. You do not need the negative sign though becaue the current flow is oriented from V+ to ground.

Answer 3

Think of a water hose. The water flowing in the hose is the current. There is water pressure pushing the current through the hose. You get faster water flow with a big hose than you get with a small hose. The hose provides resistance. Pinching the hose increases the resistance, which makes the waterflow smaller. Electricity is the same way. Current is the charged particles flowing (electrons in a wire) Pressure is the voltage pushing the charged particles through the wire. You get more current flow through a big wire than a little wire. The battery cables in your car are big wires. They can carry hundreds of amperes. The wire connecting your computer to the wall socket is a smaller wire. It can only carry about 10 to 15 amperes; if you try to push 100 amperes through it then it would get hot, melt the insulation, and even possibly melt the copper wire inside. The battery cable in your car has a low resistance, the wire to your computer has a higher resistance. You have to use large-wire cables to jump-start a car for this reason. Electrical calculations are surprisingly simple. You can measure the voltage with a voltmeter. You simply put the probes across the device you want to measure. You can measure current with an ammeter. You make the ammeter part of the circuit, just like a water meter is installed in the water line. You measure voltage, you measure current, and you can calculate the resistance by Voltage divided by current = resistance. This works for battery operated equipment and it works for devices you plug into the wall sockets, but as the voltage in a wall socket is lethal, don't do this unless someone with electrician or electrical engineer skills shows you how to do it. ===== Everything has resistance at normal temperatures, even empty space; the physicists call it the characteristic impedance of a vacuum. Superconductors have no resistance, but incredibly cold temperatures are needed to create a superconductor and do something useful with it. Too expensive for home use.

Answer 4

The current is correct. i = v/r = 0.3 A
V(R1) = i * R1 = .3 * 10 = 3V
V(R2) = i * R2 = .3 * 30 = 9V

When inspecting a circuit with resistors, you can get a negative voltage reading on a multi-meter, but this just means that your leads (positive and negative) are reversed.

Answer 5

You have everything correct except for adding the negative signs.

Technically you can have a negative voltage but that is more in the direction of current flow. In this instance the voltages should be considered positive.

Answer 6

Looks good and unless your teacher insists, the - sign on the voltage drops would never be used in the real world.