Three resistors of 6Ω, 12Ω, and 36Ω, are connected in parallel and then connected in series to a fourth resistor of unknown value. The supply voltage is 72V and the current passing through the parallel branch is 6 A. Calculate the current through the other three resistors, the total current, the value of the unknown resistor, and the total resistance.

 

What current do you think is passing through the fourth resistor? Do you know how to calculate the value of resistors in parallel? It’s important to understand this and I doubt anyone will give you the answer without seeing your attempt first.

 

Solutions to such problem are based on simplification. The three resistors can be replace by a singular resistor. Then this single resistor is in series with the unknown resistor. You know the total voltage is 72V and the current is 6A. Then the total effective resistance is the voltage divided by the current. Subtract the equivalent resistance of 6Ω, 12Ω, and 36Ω in parallel from the total resistance and that tells you the unknown resistance. Verify by calculating voltage drops across the various resistances and summing voltages and currents.

 

Do you know how to calculate the equivalent resistance of parallel resistors?

 

What current do you think is passing through the fourth resistor? Do you know how to calculate the value of resistors in parallel? It’s important to understand this and I doubt anyone will give you the answer without seeing your attempt first.

Hello, Good day, knowing my known values of the 3 Resistors and the Voltage supply. I found the current for the individual resistors in parallel. That was 72V/6, 72V/12, 72V/36. I then added up each individual current to get the total current of the parallel part. After that, I calculated the Resistance Total for the parallel which was (1/12, 1/6, 1/36) then I did the reciprocal of the answer, so I can have a series equivalent to the unknown resistor that is in series. But after doing so I'm lost with the 6A flowing through the parallel branch.

 

I found the current for the individual resistors in parallel. That was 72V/6, 72V/12, 72V/36.

How can there be 72V across the parallel resistors when they are in series with another resistor?

 

How can there be 72V across the parallel resistors when they are in series with another resistor?

This is how I have it drawn out, thus I had the 72V = for the 3 knowns.

 

You’ve drawn all four resistors in parallel

 

You’ve drawn all four resistors in parallel

Seriously?

 

Hello,

Have a look at the eBook:
https://www.allaboutcircuits.com/te...chpt-5/what-are-series-and-parallel-circuits/

Bertus

 

Seriously?

Yes

 

Seriously?

Would we kid you about that?

The resistor in series needs to carry the current of all the resistors in parallel.

 

Would we kid you about that?

The resistor in series needs to carry the current of all the resistors in parallel.

Ahhh, I see. It just came as a surprise as I've been drawing circuits like this for 2 weeks now and wasn't told anything about how it's structured.

 

Ahhh, I see. It just came as a surprise as I've been drawing circuits like this for 2 weeks now and wasn't told anything about how it's structured.

So let's see your schematic of the correct circuit.

 

So let's see your schematic of the correct circuit.

Is this correct or still wrong?

 

Is this correct or still wrong?
View attachment 279260

That's correct.
Now can you calculate the equivalent parallel value and the unknown series resistor value?

 

That's correct.
Now can you calculate the equivalent parallel value and the unknown series resistor value?

ohhhh yeah, now it's clearer what to do. Thank you very much for the correct guidance. I'll take some time out to perfect circuit schematics, to minimize my confusion.

 

Hello, Good day, knowing my known values of the 3 Resistors and the Voltage supply. I found the current for the individual resistors in parallel. That was 72V/6, 72V/12, 72V/36. I then added up each individual current to get the total current of the parallel part. After that, I calculated the Resistance Total for the parallel which was (1/12, 1/6, 1/36) then I did the reciprocal of the answer, so I can have a series equivalent to the unknown resistor that is in series. But after doing so I'm lost with the 6A flowing through the parallel branch.

You are making one of the most common mistakes among people new to Ohm's Law. You are picking any voltage and any resistance and throwing them at Ohm's Law to get a current (or, in general, picking two values to get the third) without considering whether Ohm's Law even applies to those two values.

Ohm's Law is very specific. It relates a resistance to the current flowing through THAT resistance and the voltage that appears across THAT resistance. While all three parallel resistors do have the same voltage across them (that's what makes them be in parallel), that voltage is NOT the 72 V, which is the supply voltage that is applies across the combination of all four resistors.

 

View attachment 279249View attachment 279249This is how I have it drawn out, thus I had the 72V = for the 3 knowns.

The thing to keep in mind are the following definitions for series and parallel:

Two (two-terminal) components are in series if and only if whatever current flows in one component MUST flow in the other.

Two (two-terminal) components are in parallel if and only if whatever voltage appears across one component MUST appear across the other.

An easy check for parallel components is to label or color code the nodes.



All components that are connected to both Node A (the red node) and Node B (the blue node), are in parallel.

The notion of series and parallel get murkier when components have more than two terminals.

Also, as you will discover, you can hook up nothing but two terminal components (say resistors) in such a way that no two components are in series and no two components are in parallel. That's another mistake that people commonly make at this stage -- getting it in their heads that every circuit consists of some combination of series and parallel components.