voltage drop

[Guest larasmith]

i was just wondering about this concept. why does voltage drop add in a series yet stay the same in parallel circuit? and what exactly do they mean by voltage DROP?

[Guest skofu]

Voltage differences in circuits follow Ohm's law: (delta) V = IR. How large the voltage drop across a given set of resistors in a circuit is will depend on how everything is connected in the circuit, and on the type of power source.

 

Some examples:

 

Consider a voltage source providing a voltage V0 connected to a number of resistors (let's say 3). The voltage source provides a constant voltage of V0 across the entire circuit. the total current in the circuit and the voltage drop across each individual resistor are the things that will change depending on whether the resistors are arranged in parallel or series. If the resistors are in series, then the total resistance (Rt) is the linear sum of each individual resistance: Rt = R1+R2+R3. So the current flowing in this circuit is V0/Rt, from ohm's law. for the sake of simplicity, let's say that R1=R2=R3=R (where R can be any arbitrary resistance - 10 Ohms for example). In this case, the total current is V0/(3R) (ohm's law). The voltage drop across each resistor (Vr) will be Vr = IR = [V0/(3R)]*R = V0/3. If we add another resistor, the total current becomes V0/(4R) and the individual voltage drop across each resistor will be V0/4, etc. No matter how many resistors we add in series, the total voltage drop in the entire circuit remains V0, but the current and voltage drop across each resistor will change.

 

If the resistors are in parallel, the voltage drop across each resistor will be V0, because we have a voltage source supplying a constant voltage, and in this case the voltage drop across any given resistor is equal to the voltage drop in the entire circuit - draw a circuit diagram if you need to convince yourself of this. The current in each resistor will now be V0/R, since the voltage drop across each resistor must obey ohm's law. That means that for each resistor, an amount of current equal to V0/R must flow from the power supply. In this case, the voltage drop in the circuit and in each resistor stays the same no matter how many resistors we add in parallel, as you describe, but the current coming out of the power source will increase by V0/R for every resistors we add.

 

Now consider a current power source connected to the resistors. It supplies a constant current I0. If the resistors are in parallel, the current must split between them. If we have R1=R2=R3=R, the current divides evenly between the 3 resistors, and the voltage drop across any of the resistors will be I0*R/3, since the current in each resistor is I0/3. In this case, if we add resistors in parallel, we will decrease the current flowing in each individual resistor (because the current being supplied is constant, and it will be divided among more resistors in parallel as we add them) Thus the voltage drop in the circuit will decrease as we add resistors.

 

If the resistors are in series with a current power source, then the total resistance is Rt = R1+R2+R3. Since we are supplying a constant current of I0 which is NOT being divided as it was in the parallel circuit, but is flowing through all of the resistors, we will have V = I0*Rt. If we add resistors in this case, the voltage drop across the entire circuit will increase, because we are forcing the same current (I0) across a larger and larger resistance, as you describe.

 

Hope this helps,

s.

[Guest larasmith]

thanks so much, skofu!! that really helped!

[Guest skofu]

no prob!

 

s.