# Finding the resistance in a parallel-series circuit.

• Eclair_de_XII
In summary, we are given a circuit with an emf of 24V and resistors R1, R2, and R3. By applying KCL and KVL, we can find the values of R2 and R3, assuming negligible resistance. Using the equation for voltage in parallel circuits, we find R2 to be 2.265 V/A. Similarly, using the equation for total voltage, we find R3 to be 1.7 V/A. However, in a later conversation, we find that the last line of the solution is incorrect and that the sum of potential differences along a closed loop is 0 (KCL). We also discuss the concept of potential differences and how they add up along a loop
Eclair_de_XII

## Homework Statement

"Find ##R_2## and ##R_3##, assuming negligible resistance ##r=0\frac{V}{A}##."

##R_1=6.1\frac{V}{A}##
##I_1=1.3A##
##I_3=4.8A##
##\epsilon = 24 V##

## Homework Equations

Voltage in parallel circuits: ##V_1=V_2## or ##I_1R_1=I_2R_2##
Total voltage: ##\epsilon = I_1R_1+I_2R_2+I_3R_3##

## The Attempt at a Solution

##I_3=I_1+I_2##
##I_2=I_3-I_1=3.5 A##

##R_2=\frac{I_1}{I_2}R_1=2.265\frac{V}{A}##

##R_3=\frac{1}{I_3}(\epsilon -I_1R_1+I_2R_2)=1.7\frac{V}{A}##

I'm new to circuits, so if anyone would like to care to tell me where I went wrong, that would be much appreciated. Thank you.

Eclair_de_XII said:

## Homework Statement

"Find ##R_2## and ##R_3##, assuming negligible resistance ##r=0\frac{V}{A}##."
View attachment 212066
##R_1=6.1\frac{V}{A}##
##I_1=1.3A##
##I_3=4.8A##
##\epsilon = 24 V##

## Homework Equations

Voltage in parallel circuits: ##V_1=V_2## or ##I_1R_1=I_2R_2##
Total voltage: ##\epsilon = I_1R_1+I_2R_2+I_3R_3##
The last line is wrong. It would be true if all resistors were connected in series.
Eclair_de_XII said:

## The Attempt at a Solution

##I_3=I_1+I_2##
##I_2=I_3-I_1=3.5 A##

##R_2=\frac{I_1}{I_2}R_1=2.265\frac{V}{A}##

##R_3=\frac{1}{I_3}(\epsilon -I_1R_1+I_2R_2)=1.7\frac{V}{A}##
Wrong. What is the voltage across R3?

It's not ##V_3=I_3R_3##? In any case, could I not replace the two parallel resistors by an equivalent resistor so that everything is in series?

##\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_2+R_1}{R_1R_2}##

So ##R_{eq}=\frac{R_1R_2}{R_2+R_1}##. And now we have...

##\epsilon = I_3(\frac{R_1R_2}{R_2+R_1}+R_3)## or ##R_3=\frac{\epsilon}{I_3}-\frac{R_1R_2}{R_2+R_1}##

In any case, I do not know what ##V_3## is supposed to be, but I solved the problem. Thanks for telling me about my error at the very end. But just so I know, what would the sum of voltages look like for a parallel circuit look like?

Eclair_de_XII said:
In any case, I do not know what ##V_3## is supposed to be, but I solved the problem. Thanks for telling me about my error at the very end. But just so I know, what would the sum of voltages look like for a parallel circuit look like?
The sum of the potential differences along a closed loop is 0 (KCL) With respect to point O, UA=E, UB= ? UC=:?

ehild said:
With respect to point O, UA=E, UB= ? UC=:?

I think ##U_B=-E##?

Eclair_de_XII said:
I think ##U_B=-E##?
Why?
(Also, why have you marked this thread solved?)

cnh1995 said:
Why?
(Also, why have you marked this thread solved?)

I think it's because ##U_A+U_B=0##. I marked it solved because I figured it out by replacing the two parallel resistors with an equivalent resistor, then using that for my voltage equation for series resistors (see post #3). But since I still need help, should I mark it unsolved?

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Eclair_de_XII said:
I think it's because UA+UB=0UA+UB=0U_A+U_B=0.
ehild said:
The sum of the potential differences along a closed loop is 0 (KCL) With respect to point O, UA=E, UB= ? UC=:?
Eclair_de_XII said:
But since I still need help, should I mark it unsolved?
Oh I see you've solved your original question and need some additional help. So no need to mark it unsolved.

##\epsilon - (U_A+U_B+U_C)=0V## is all I'm gathering from this. Also, I don't really know where ##C## is defined to be.

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Eclair_de_XII said:
##\epsilon - (U_A+U_B+U_C)=0V## is all I'm gathering from this. Also, I don't really know where ##C## is defined to be.
Sorry, I forgot to show point C.

I asked the potentials at A, B, C, with respect to point O. Can you calculate the numerical value of UB? What is UC then? What is the potential difference across R3 ?

I guess I'm going to assume that the parallel circuit is the closed loop...? So ##-U_A-U_B+U_A+\epsilon=0## or ##U_B=\epsilon##...?

Eclair_de_XII said:
I guess I'm going to assume that the parallel circuit is the closed loop...? So ##-U_A-U_B+U_A+\epsilon=0## or ##U_B=\epsilon##...?
You confuse potential with potential difference. UA, UB are potentials at point A and B. KVL states that the sum of potential differences (voltages) in a closed loop is zero.
What is the potential difference UA-UB across R1 if R1= 6.1 Ω and I1=1.6 A?

ehild said:
What is the potential difference UA-UB across R1 if R1= 6.1 Ω and I1=1.6 A?

Oh, that's just ##V_1=V_2=(R_1)(I_1)=(6.1Ω)(1.6 A)=9.76 V##, I think.

Eclair_de_XII said:
Oh, that's just ##V_1=V_2=(R_1)(I_1)=(6.1Ω)(1.6 A)=9.76 V##, I think.
Sorry, I1 was 1.3 instead of 1.6. So the voltage across both R1 and R2 is 7.93 V.
What is the direction of the current through R1?

Well, it goes from positive to negative, so the negative direction? Wait, it starts from ##0V## then goes to ##7.93V##, so the positive direction.

So relative to ##O##, ##U_A=0V##, and ##U_B=U_C=7.93V##, some point after the third resistor is ##U_3=16.07V##, and at the emf, ##U_{emf}=24V##, I think.

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Eclair_de_XII said:
Well, it goes from positive to negative, so the negative direction?
Yes, the current flows from A to B.

Point A is connected to the positive terminal of the battery, it is at 24 V with respect to the negative terminal. The potential drops across R1 in the direction of the current. If A is at 24 V and the potential drops by 7.93 V across R1, what is the potential at point B?
What is the potential difference across R3?

Let's see... ##U_B=16.07V##, and I guess ##U_C=7.93V##. So ##U_A-U_B-U_C=0##?

Eclair_de_XII said:
Let's see... ##U_B=16.07V##, and I guess ##U_C=7.93V##. So ##U_A-U_B-U_C=0##?
Yes, UB=16.07 V with respect to O. B and C are connected by a wire of zero resistance, so they are at the same potential: UC=16.07 V. What is the potential difference across R3?
Try to understand that UA, UB, UC are potential differences. NO sense to add them. The potential differences along a loop add up to zero, not the potentials.
Imagine you are at zero high, at a seaside. Then you go up a hilltop that is 500 m high. Then you descend 100 m to a cave, at 400 m height. From here you go down 200 m and you arrive at a camp, at height of 200 m. From there you go down 200 m and you reach the seaside, at 0 height. . The high differences of your walk add up to zero, but not he heights themselves: 500-100-200-200 = 0, but 500+400+200 is not zero!

## What is a parallel-series circuit?

A parallel-series circuit is a combination of both parallel and series components. This means that the circuit has branches that are connected in parallel, but those branches are then connected in series with each other.

## How is the resistance calculated in a parallel-series circuit?

The total resistance in a parallel-series circuit is calculated by adding the resistance of all the branches together. This can be done using the formula R = R1 + R2 + R3... where R is the total resistance and R1, R2, R3 are the resistances of each branch.

## What is the difference between series and parallel resistors?

In a series circuit, the resistors are connected one after the other, creating only one path for the current to flow. In a parallel circuit, the resistors are connected side by side, creating multiple paths for the current to flow. This results in a different total resistance for each type of circuit.

## Can a parallel-series circuit have a total resistance of 0?

No, a parallel-series circuit cannot have a total resistance of 0. This is because there will always be some resistance in each branch, and when added together, the total resistance will never be 0.

## How can I measure the resistance in a parallel-series circuit?

The resistance in a parallel-series circuit can be measured using a multimeter. Set the multimeter to the resistance (ohms) setting and connect the probes to each end of the circuit. The reading on the multimeter will be the total resistance of the circuit.

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