Find Potential Difference between X & Y in Circuits with Parallel & Series Arr.

[Lim Y K]

Homework Statement

In a circuit consisting of both parallel and series arrangement, how do you find the potential difference of X(in red) and Y(in blue)? and how did you derive at those answers?

Homework Equations

idk

The Attempt at a Solution

total resistance=4 ohms +(1/r eff=1/2ohms+1/4ohms)
=4ohms+4/3ohms
=5 1/3 ohms
current=16 divide by 5 1/3ohms
=3 ampere
potential diff for X ------V=RI
=(1/2)(3)
=1.5
potential diff for Y-------V=RI
=(1/4)(3)
= 3/4

I WAS TOLD THOSE IN RED ARE WRONG. BUT WHY? I DIDN'T SEE ANY MISTAKE IN CALCULATION.
can someone help me please? please explain to me what went wrong thanks

 

[DEvens]

If 3 Amps is flowing through the battery, how much is flowing through each resistor in the circuit? Then what does V=RI tell you for each resistor?

If you know the potential drop across the "other" resistor, then what does that tell you about the potential drop across the combination of X and Y? What total do you have to make?

 

[Lim Y K]

If 3 Amps is flowing through the battery, how much is flowing through each resistor in the circuit? Then what does V=RI tell you for each resistor?

If you know the potential drop across the "other" resistor, then what does that tell you about the potential drop across the combination of X and Y? What total do you have to make?

according to i1=i2=i3, 3 amps flowing through batt means 3 amps flowing through the resistors. I=v/R means 3=v/ 1/2 for resistor x and 3= v/ 1/4 for resistor y

potential drop in other resistor= R/I
=4/3
i know of the V1=V2=V3 rule but why is 4/3 volts= potential diff of both X and Y?

 

[andrevdh]

Electric current behaves much like water flowing in a river.
That means that part of the 3 A flows through X and the other part through Y,
but not necessarily the same amount through each. Less will go through Y and
more through X, but their sum will be 3A. Since V = IR I do not understand why
you are inverting the resistance?

 

[Dembadon]

according to i1=i2=i3, 3 amps flowing through batt means 3 amps flowing through the resistors. I=v/R means 3=v/ 1/2 for resistor x and 3= v/ 1/4 for resistor y

potential drop in other resistor= R/I
=4/3
i know of the V1=V2=V3 rule but why is 4/3 volts= potential diff of both X and Y?

I'm not sure where you are getting those current and voltage relationships. Purely resistive circuits can have currents and voltages that equal each other, yes, but only under certain element configurations. You should review the conditions under which such relationships are valid. Now, on to your problem:

You need to find total current. You were on the right track with using Ohm's Law and the equivalent resistance of the circuit, but you should not be inverting the resistances.

You can then use current division to solve for the current in the resistors.

Here is the general form for current division:

I_x = R_y*I_Total / (R_x + R_y), where I_Total is the current you've already found, and I_x is the current through resistor x.

After you have found the current through each resistor, you can use Ohm's Law to find the voltage drop across each.

 

[andrevdh]

A parallel combination of resistors, R₁, R₂, R₃... can be replaced by a single resistor R_P
where R_P should be

1/R_P = 1/R₁ + 1/R₂ + 1/R₃ ...

this means that a resistor R₁ in a parallel circuit can be replaced by a resistor R_P where

1/R_P = 1/R₁

so that R_P= R₁ and not

R_P = 1/R₁

 

[Dembadon]

A parallel combination of resistors, R₁, R₂, R₃... can be replaced by a single resistor R_P
where R_P should be

1/R_P = 1/R₁ + 1/R₂ + 1/R₃ ...

this means that a resistor R₁ in a parallel circuit can be replaced by a resistor R_P where

1/R_P = 1/R₁

so that R_P= R₁ and not

R_P = 1/R₁

In this case, since there are only two resistors in parallel, one can use the following relationship:

R_Equiv = R_x*R_y / (R_x + R_y)

But yes, when one wants to combine more than two parallel resistances, the methods you've described is required.