Finding the resistance in a parallel-series circuit.

[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.

 

[ehild]

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.

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?

 

[Eclair_de_XII]

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?

 

[ehild]

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=:?

 

[Eclair_de_XII]

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

I think $U_B=-E$?

 

[cnh1995]

I think $U_B=-E$?

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

 

[Eclair_de_XII]

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?

 

[cnh1995]

I think it's because UA+UB=0UA+UB=0U_A+U_B=0.

No. Read ehild's post again.

The sum of the potential differences along a closed loop is 0 (KCL) With respect to point O, UA=E, UB= ? UC=:?

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.

 

[Eclair_de_XII]

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

 

[ehild]

$\epsilon - (U_A+U_B+U_C)=0V$ 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 ?

 

[Eclair_de_XII]

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 $U_B=\epsilon$...?

 

[ehild]

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?

 

[Eclair_de_XII]

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.

 

[ehild]

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?

 

[Eclair_de_XII]

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.

 

[ehild]

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?

 

[Eclair_de_XII]

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

 

[ehild]

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!