Calculating Voltage and Current in a Series Circuit

[Adeel Ahmad]

Homework Statement

Find the voltage v and the currents i1 and i2 for the circuit below.

Homework Equations

V=IR

The Attempt at a Solution

I would combine the 10 ohm and 40 ohm, and the 18 ohm and 9 ohm, so that they are in series with the 6 ohm resistance. Then I would use V=IR to get the current, but not sure if this is the correct approach.

 

[gneill]

Homework Statement

Find the voltage v and the currents i1 and i2 for the circuit below.

Homework Equations

V=IR

The Attempt at a Solution

I would combine the 10 ohm and 40 ohm, and the 18 ohm and 9 ohm, so that they are in series with the 6 ohm resistance. Then I would use V=IR to get the current, but not sure if this is the correct approach.

I think you mean you could use that method to find the voltage v across the 5 A current supply. Yes, that would work.

You're still left with finding the individual currents $i_2$ and $i_2$. You know about voltage dividers (from a previous thread), but do you also know about current dividers?

 

[Adeel Ahmad]

I think you mean you could use that method to find the voltage v across the 5 A current supply. Yes, that would work.

You're still left with finding the individual currents $i_2$ and $i_2$. You know about voltage dividers (from a previous thread), but do you also know about current dividers?

View attachment 106125

No, I am not aware of current dividers

 

[Adeel Ahmad]

I think you mean you could use that method to find the voltage v across the 5 A current supply. Yes, that would work.

You're still left with finding the individual currents $i_2$ and $i_2$. You know about voltage dividers (from a previous thread), but do you also know about current dividers?

View attachment 106125

So I looked it up, and this is what I calculated. I made all resistances in series and added them and got 20 ohms. V=IR = (2A)(20 ohms) = 40V. Then i1 = i_totalR_total / R1 = (2A)(20 ohms) / (12 ohms) = 3.3A

 

[Adeel Ahmad]

i2 will also be the same as i1 according to my calculations since resistances across each will be equal

 

[Adeel Ahmad]

I am using different values then the ones show in the picture, they are 2A for the total current instead of 5A, and from left to right, the resistances are 12, 24, 8, 6, and 12

 

[gneill]

So I looked it up, and this is what I calculated. I made all resistances in series and added them and got 20 ohms. V=IR = (2A)(20 ohms) = 40V. Then i1 = i_totalR_total / R1 = (2A)(20 ohms) / (12 ohms) = 3.3A

A current can't be larger than the current source. So your 3.3 A result should send up a flag that something's off.

Your problem is that you've got the current divider equation a bit mixed up. You want to multiply the total current by the resistance of the "other" path divided by the total resistance. So for example, using the figure I posted in post #2,
$$i_1 = I \frac{R_2}{R_1 + R_2}$$
Note that the "other" path in this case is comprised of $R_2$.

 

[Adeel Ahmad]

A current can't be larger than the current source. So your 3.3 A result should send up a flag that something's off.

Your problem is that you've got the current divider equation a bit mixed up. You want to multiply the total current by the resistance of the "other" path divided by the total resistance. So for example, using the figure I posted in post #2,
$$i_1 = I \frac{R_2}{R_1 + R_2}$$
Note that the "other" path in this case is comprised of $R_2$.

Therefore, i1 = (2A)(12 ohms) / (12 ohms + 12ohms) = 1 amp
Since R1 and R2 are equal, i1 and i2 will also have the same value.
Would that be correct?

 

[epenguin]

Nothing wrong with what you said and did – but it's a bit pedestrian and following textbook routine.
You should be able to say the currents off the top of your head!
That is what gneil is getting at saying "current dividers".
Certainly off the top of your head you can say immediately what the current is in the 6 Ω resistor. OK you're asked what it is in the branch parts. Current flows the parallel conductors just in proportion to their conductances.

 

[gneill]

Therefore, i1 = (2A)(12 ohms) / (12 ohms + 12ohms) = 1 amp
Since R1 and R2 are equal, i1 and i2 will also have the same value.
Would that be correct?

No, There are two separate current dividers to consider: one divider consists of a 12 Ω and a 24 Ω resistor while the other consists of a 6 Ω and a 12 Ω resistor. See the image below. One current division is shown in blue, the other green.

 

[epenguin]

Maybe repeating, but you should find this really easy.

Referring for instance to your original #1 values, Just because the arithmetic is particularly simple, for the left-hand branch 5A enters and leaves, dividing itself between conductances which are in the ratio 4:1. the current distributes itself in that ratio.