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Resistor currect and voltage drop in a circuit.

by davidepalmer
Tags: circuit, currect, resistor, voltage
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davidepalmer
#1
Feb22-10, 03:20 PM
P: 8
Consider this circuit with R1 = 30 ohm, R2 = 10 ohm, R3 = 10 ohm, R4 = 20 ohm,
R5 = 60 ohm. (the circuit is attached)

A. What is the current across the resistor R2, if a voltage of 110 Volt is applied to
the two terminals at the bottom of the figure?
B. What is the voltage drop across R2?
C. Keep all other resistances at the same value, but allow R2 to vary. What are the
minimum and maximum resistances that this circuit can have?


I know that i have to use V=IR to solve for the voltage drop and the current across the resistor.

I have already found the equivilent resistance of the arrangement. I am able to find the total current throughout the whole circuit, but after that I don't really know where to start.
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xcvxcvvc
#2
Feb22-10, 03:27 PM
P: 393
Quote Quote by davidepalmer View Post
Consider this circuit with R1 = 30 ohm, R2 = 10 ohm, R3 = 10 ohm, R4 = 20 ohm,
R5 = 60 ohm. (the circuit is attached)

A. What is the current across the resistor R2, if a voltage of 110 Volt is applied to
the two terminals at the bottom of the figure?
B. What is the voltage drop across R2?
C. Keep all other resistances at the same value, but allow R2 to vary. What are the
minimum and maximum resistances that this circuit can have?


I know that i have to use V=IR to solve for the voltage drop and the current across the resistor.

I have already found the equivilent resistance of the arrangement. I am able to find the total current throughout the whole circuit, but after that I don't really know where to start.
After you have the total current through one jumbo resistor [tex] R_{eq}[/tex] you can begin to then reverse your resistor simplifications one step at a time to find the voltage and current for each resistor.

Use these two rules: the voltage across parallel resistors is the same and the current through series resistors is the same. Therefore, let's say you unsimplified your jumbo resistor into two parallel resistors. You know that the voltage across that 1 jumbo resistor is across both those parallel resistors. If you had expanded it to two series elements, you know the current through each of them is the same as the current through the jumbo resistor. So you keep stepping back and using these rules until you have everything you need.
davidepalmer
#3
Feb22-10, 03:35 PM
P: 8
Quote Quote by xcvxcvvc View Post
After you have the total current through one jumbo resistor [tex] R_{eq}[/tex] you can begin to then reverse your resistor simplifications one step at a time to find the voltage and current for each resistor.

Use these two rules: the voltage across parallel resistors is the same and the current through series resistors is the same. Therefore, let's say you unsimplified your jumbo resistor into two parallel resistors. You know that the voltage across that 1 jumbo resistor is across both those parallel resistors. If you had expanded it to two series elements, you know the current through each of them is the same as the current through the jumbo resistor. So you keep stepping back and using these rules until you have everything you need.
Therefore if i'm able to find the voltage after R1, that voltage is the same through R4 and R23. So with this information I can find the current through R23 and since R2 and R3 are in series they have the same current as R23?

xcvxcvvc
#4
Feb22-10, 04:09 PM
P: 393
Resistor currect and voltage drop in a circuit.

Quote Quote by davidepalmer View Post
Therefore if i'm able to find the voltage after R1, that voltage is the same through R4 and R23. So with this information I can find the current through R23 and since R2 and R3 are in series they have the same current as R23?
I'm not sure what you mean by "find the voltage after R1"

[tex]R_{eq} = R_1 + R_4||(R_2+R_3) +R_5[/tex]
This is how you simplified the circuit into 1 resistor. Here is each step shown visually:


So here you are with a current through the 1 resistor in step 3. Let's unsimplify the circuit by 1 step, and go back to step 2. As you can see, there are three series resistors. We know that the current through the series resistors is the same as the current of the one equivalent circuit -- where else would the current flow since the resistors are one a one way track? So with whatever that current is you found, use Ohm's law to find the voltage across the resistor that is or that contains the resistor we want to know about. In this example, the middle resistor contains R2, the resistor we want.

With the voltage across [tex]R_4||(R_2+R_3)[/tex] found by [tex] I *( R_4||(R_2+R_3))[/tex], we are ready for the next step: step back in the simplification again. We are now at picture 1. We see then that the voltage we JUST found across [tex]R_4||(R_2+R_3)[/tex] is the same voltage across [tex] R_2 + R_3[/tex]. Find the current through that 1 equivalent resistor then step it back to 2 series elements R_2 and R_3. We then use the current rule that current is the same through series elements so we know the current through R_2 as well. Ohms law then gives the voltage across R_2.
davidepalmer
#5
Feb22-10, 04:32 PM
P: 8
Quote Quote by xcvxcvvc View Post
I'm not sure what you mean by "find the voltage after R1"

[tex]R_{eq} = R_1 + R_4||(R_2+R_3) +R_5[/tex]
This is how you simplified the circuit into 1 resistor. Here is each step shown visually:


So here you are with a current through the 1 resistor in step 3. Let's unsimplify the circuit by 1 step, and go back to step 2. As you can see, there are three series resistors. We know that the current through the series resistors is the same as the current of the one equivalent circuit -- where else would the current flow since the resistors are one a one way track? So with whatever that current is you found, use Ohm's law to find the voltage across the resistor that is or that contains the resistor we want to know about. In this example, the middle resistor contains R2, the resistor we want.

With the voltage across [tex]R_4||(R_2+R_3)[/tex] found by [tex] I *( R_4||(R_2+R_3))[/tex], we are ready for the next step: step back in the simplification again. We are now at picture 1. We see then that the voltage we JUST found across [tex]R_4||(R_2+R_3)[/tex] is the same voltage across [tex] R_2 + R_3[/tex]. Find the current through that 1 equivalent resistor then step it back to 2 series elements R_2 and R_3. We then use the current rule that current is the same through series elements so we know the current through R_2 as well. Ohms law then gives the voltage across R_2.
I think I've finally figured out what you've been explaining to me. I was able to work it backwards and get the answer now. Thank you so so much for your help!


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