# Why cannt I calculate the voltage of the second resistor analog to the first resistor?

#### arhzz1

Joined Oct 21, 2020
42
Hello!

I have a question regarding this circuit;

Now the values of U = 10V and R1 = 150 Ohm R2 = 470 Ohm and R3 = 330 Ohm.

Consider the case that S1 closes the circuit and the current flows through the resistor R3 as well. I was calculating the voltage at R3 and I did so successfully using the Kirchoffs Loop Law.The voltage should be U3 = 5,65V. But here is the Problem that was bothering me.

I first tried doing it like this. I new that R1 and R2 were in series so I calculated that the current has to be I = U/R1+R2; since R3 and R2 are parallel they should have the same voltage;hence calculating U2 would give me U3.So I tried U2 = I * R2 and I got the wrong result.Than I kept trying and trying and I solved it so that I calculated U1 = I * R1 and than I used 2 Loops to get the value of U2 and it turns out U2 was equal to U3.

Now my question is why didnt my first method work? If I was able to calculate U1 = R1 * I why didnt it work for U2 = R2 * I when the same current is flowing through both of them?

I know that the answer is probably trivial to most of you but that I am just starting out this is very unlogical to me and an explanation would help.

Thanks!

#### ericgibbs

Joined Jan 29, 2010
14,712
I first tried doing it like this. I new that R1 and R2 were in series so I calculated that the current has to be I = U/R1+R2; since R3 and R2 are parallel they should have the same voltage
hi arh,
Why do you conclude that.?

Hint: what effect would be connecting R3 in || R2 have on the current thru R1

E

#### arhzz1

Joined Oct 21, 2020
42
hi arh,
Why do you conclude that.?

Hint: what effect would be connecting R3 in || R2 have on the current thru R1

E
I am a bit confused with your question but I'll give my best.

Why do I conclude that the voltage in R2 and R3 has to be the same? Well since they are parallel to each other right? And for your hint; I am guessing you are asking if I would to calculate the resistance of R3 || R2 what effect would it have on the current flowing through R1? well I'd expect that the current flowing through R1 would be divided?

#### ericgibbs

Joined Jan 29, 2010
14,712
hi arh,
Do this little bit of maths..
Calculate and post the voltage at the junction of R1 and R2 , without R3 being connected.
Post the value.

Calc the parallel resistive value of R2 and R3, in parallel call it R4,
relcalc the junction voltage of R1 and R4, post what you calculate

E

#### arhzz1

Joined Oct 21, 2020
42
hi arh,
Do this little bit of maths..
Calculate and post the voltage at the junction of R1 and R2 , without R3 being connected.
Post the value.

Calc the parallel resistive value of R2 and R3, in parallel call it R4,
relcalc the junction voltage of R1 and R4, post what you calculate

E
I dont know how to calculate the voltagw of a junction, we have not covered that in class.

#### ericgibbs

Joined Jan 29, 2010
14,712
hi
OK,
Have you covered Ohms Law.?
V= I * R, I =V/R, R=V/I

E

#### arhzz1

Joined Oct 21, 2020
42
hi
OK,
Have you covered Ohms Law.?
V= I * R, I =V/R, R=V/I

E
Yes I am familiar with those

#### ericgibbs

Joined Jan 29, 2010
14,712
OK.
Do it by steps if you wish.

1. What is the total resistance of R1 and R2 in Series. ?

2, What is the resistance of R2 and R3 in parallel.? [ Call the new value say R4]

3. For step #1 what is the current through R1 and R2 in series.?

4. What is the total resistance of R1 and R4 in Series. ?

5. What is the current through R1 and R4 in series.?

E
typo**

#### arhzz1

Joined Oct 21, 2020
42
Okay step by step

1. R1+R2 = 620 Ohm

2.R4 =R2 || R3 = 193,875 Ohm

3.I = U/R1+R2 = 0,016 A

4. R1+R4 = 343,875 Ohm

5. I = U/R4 = 0,029 A

#### ericgibbs

Joined Jan 29, 2010
14,712
hi,
OK,
So you know for both cases the current through R1.

So what is the voltage drop across R1 in both cases.
Subtract that voltage from 10V , for both cases and that will give you two voltage values at the junction
So
V1= 10v - VdropR1= I * R1 = ................
V2 =10v - VdropR1 =I * R1 = .........

E

BTW: I would advise you to use a suffix to identify voltage and current values in your equations
I1 = U/R1+R2 = 0,016 A
I2 = U/R4 = 0,029 A

V1= 10v - VdropR1= I1 * R1 = ................
V2 =10v - VdropR1 =I2 * R1 = .........

Last edited:

#### arhzz1

Joined Oct 21, 2020
42
Okay noted.

So the voltages at the junctions are

U1 = 10- 0,016 * 150 = 7,6V
U2 = 10- 0,029*150 = 5,65V

Also, for all of the readers the notation U1/V1 represents the voltage just the notation is different (American to European)

#### dl324

Joined Mar 30, 2015
13,846
So the voltages at the junctions are
You call them junctions instead of nodes? What do you call nodal equations?

#### ericgibbs

Joined Jan 29, 2010
14,712
hi arh,
So you have now proved that the junction of R1 and R2 and the junction of R1 and R2||R3 are in fact different voltage levels.

Nicely done.
E

BTW:
When I refer to junction, it is the connection point of two physical components.

#### arhzz1

Joined Oct 21, 2020
42
hi arh,
So you have now proved that the junction of R1 and R2 and the junction of R1 and R2||R3 are in fact different voltage levels.

Nicely done.
E

BTW:
When I refer to junction, it is the connection point of two physical components.
Okay so their voltages are diffrent, is that why I cannot calculate the voltage at R2 the same way I calculate it at R1?

#### ericgibbs

Joined Jan 29, 2010
14,712
hi,
As you can see, the total current thru R1 is different for the two cases, so it is the voltage drop across R1 at these two different currents which give a different voltage at the junction.

Consider that R2 was removed and you just had R1 and R 3 in series, again the current would be a different value, because of the different total resistance across the 10v supply, so the voltage drop across R1 would be different.

E

#### arhzz1

Joined Oct 21, 2020
42
hi,
As you can see, the total current thru R1 is different for the two cases, so it is the voltage drop across R1 at these two different currents which give a different voltage at the junction.

Consider that R2 was removed and you just had R1 and R 3 in series, again the current would be a different value, because of the different total resistance across the 10v supply, so the voltage drop across R1 would be different.

E
Okay so the way I understood it is. Since the currents flowing through R1 are creating a voltage drop, and the diffrent voltages at the junction are basically the voltages of R2? So I can say the voltage drop at R1 is the voltage at R2?

#### ericgibbs

Joined Jan 29, 2010
14,712
hi,
No, that would be incorrect.
You must consider the voltage driving source U=10V.
To get the voltage at the junction, you must subtract the Voltage drop across R, due to the current flowing thru R1,, from the 10V source

ie: V at junction = 10V - ( I * R1) .

#### arhzz1

Joined Oct 21, 2020
42
hi,
No, that would be incorrect.
You must consider the voltage driving source U=10V.
To get the voltage at the junction, you must subtract the Voltage drop across R, due to the current flowing thru R1,, from the 10V source

ie: V at junction = 10V - ( I * R1) .
Ah okay I see it now. Thanks a lot for your help Eric !

#### ericgibbs

Joined Jan 29, 2010
14,712
hi arh,
Just to round off.
E

I1 = U/R1+R2 = 0,016 A
I2 = U/R4 = 0,029 A

V1= 10v - VdropR1= 0.016A*150 = 10v - 2.4 = 7.6v
V2 =10v - VdropR1 =0.029 * 150 = 10v - 4.35 = 5.65v