# Parallel resistors - quick question.

#### Kango

Joined Sep 22, 2007
30
I have 2 circuits where the resistors are parallel connected but if you look at the circuit (Fig 2) it´s getting all blurry too me.

Fig 1:

R1 // R2

Fig 2:

In my opinion this should be

(R1//R2) + R3 (How can this be so?)

#### recca02

Joined Apr 2, 2007
1,211
i cant see the fig 1 and 2.
are they posted or is there a specific page u are talking about.

#### Kango

Joined Sep 22, 2007
30
i cant see the fig 1 and 2.
are they posted or is there a specific page u are talking about.
Wops, forgot to post the figures.

#### nomurphy

Joined Aug 8, 2005
567
From a DC perspective, you effectively have no parallel resistors. R1 and R2 are in series for both Fig 1 and Fig 2.

Were you to replace C in Fig 2 with a short, then you would have:
(R1||R3) + R2.

From an AC perspective of Fig 2, the math becomes a bit more complicated, because the series circuit (R3)C is in parallel with R1, which together are in series with R2.

#### bloguetronica

Joined Apr 27, 2007
1,359
I have 2 circuits where the resistors are parallel connected but if you look at the circuit (Fig 2) it´s getting all blurry too me.

Fig 1:

R1 // R2

Fig 2:

In my opinion this should be

(R1//R2) + R3 (How can this be so?)
R1 and (R3 + C) are in parallel with each other, and R2 is in series with the resulting association. It is important to not forget about the capacitors, unless the circuit is under an AC regime under a almost infinite frequency.

I cannot reply with more detail since I'm out of the context here. What do you wan't to know exacly. Are those figures taken from an exercise? If so, what is asked?

#### recca02

Joined Apr 2, 2007
1,211
until that circuit is complete it is hard to say whether even r1 will parallel r2
in fig 1 until then it is possible that the last terminals may remain open and if dc is applied they r in series.
same goes for fig 2. from what i see it is r1//r3 + r2 for DC supply.
it depends on how the circuit is closed and where is the source connected.

#### Kango

Joined Sep 22, 2007
30
Well, the question was to figure out the Frequency by "taken away "the condensator. And then using

f = 1 /2PI*Rtot*C

It seems to me that i was confusing toghter all things when taken away the capacitator, this thing is fairly basic but I mess it all up with the parallel connection.

Were you to replace C in Fig 2 with a short, then you would have:
(R1||R3) + R2.
Yes, that seems confusing to me, how come?

But wouldn´t that be (R1||R2||R3)

#### recca02

Joined Apr 2, 2007
1,211
it wud be r1//r2//r3 only if the other terminals of r1 and r2 were connected.
remember fro being in parallel the components must be connected across same two points.

#### Kango

Joined Sep 22, 2007
30
it wud be r1//r2//r3 only if the other terminals of r1 and r2 were connected.
remember fro being in parallel the components must be connected across same two points.

Thanks, I was thinking all wrong

#### nomurphy

Joined Aug 8, 2005
567
Of course, this all assumes conventional I/O with input on left and output on right. Otherwise, it's all a guessing game.

#### billbehen

Joined May 10, 2006
39
The assumption seems to be that one is operating from a low impedance source (on the left) to a high impedance load (on the right.) Thus, the output pins can be left open, but the input pins are effectively shorted! Then, from the standpoint of the cap, you have R3 in series with the other two in parallel.