Kirchhoff's 2nd Law: Terminal PD Calculation Homework

[Janiceleong26]

Homework Statement

For part (iv) 2 and 3 of this question, I don't understand why for 2) terminal p.d. of A is the e.m.f of A minus the p.d. across the internal resistance of A i.e. Va = 4.4-(0.24x2.3), I=0.24A
But for 3) the terminal p.d. is the e.m.f of B PLUS the p.d. across the internal resistance of B i.e. Vb= 2.1+(1.8x0.24)

Homework Equations

ΣE.m.f = Σp.d. around a closed circuit

The Attempt at a Solution

I minus the p.d. across the internal resistance to the e.m.f. for both cases. I can't to seem to see why is this wrong..

 

[Gmen]

I have never solved such an exercise, but since you provide the answers, I would guess the folllowing:

Cell B is connected reversely in the circuit. Thus, the pd of B is actually the voltage DROP it contributes to the circuit. Thus, its resistor actually enhances its pd.
Indeed, if you swap cell A with emf 3.848 and no resistor, and cell B with emf 2.532 and no resistor, you would get a total emf 1.316, which divided by 5.5 gives the current of 0.24A.

If you minus the pd across the internal resistor, this is wrong because it is actually one time with and one time versus the circuit loop.
What I would have done (and don't know if it would be considered wrong) is give cell B's pd as a negative value.

 

[GoodPost]

Once you apply KVL over the loop, you'll figure out the current's value and its direction is COUNTER clock wise. Now, if you want to add two series voltages up, you simply need to add the values of these voltages up, BUT you MUST consider the polarities.
In your case,
Cell A:
Since the current direction is counter clock wise, then it will pass Cell A from the right to the left. It will meet the resistance first (with positive sign)and then it will meet the negative terminal of the voltage source (negative sign).
(0.24*2.3 - 4.4 ) = - 3.848 V , the negative sign means the current will first meet the negative terminal of the new voltage.

Cell B:
The current will meet first the positive terminal of the voltage source (positive sign) and then it will find the resistance (positive sign too). As a result, we do the following
(0.24*1.8 + 2.1) = 2.532 V, similarly, the positive sign means the current will initially meet the positive terminal of the new voltage.

I hope this helps a bit.

Best regrads,
G.P.

 

[Janiceleong26]

[

I have never solved such an exercise, but since you provide the answers, I would guess the folllowing:

Cell B is connected reversely in the circuit. Thus, the pd of B is actually the voltage DROP it contributes to the circuit. Thus, its resistor actually enhances its pd.
Indeed, if you swap cell A with emf 3.848 and no resistor, and cell B with emf 2.532 and no resistor, you would get a total emf 1.316, which divided by 5.5 gives the current of 0.24A.

If you minus the pd across the internal resistor, this is wrong because it is actually one time with and one time versus the circuit loop.
What I would have done (and don't know if it would be considered wrong) is give cell B's pd as a negative value.

how do you know which internal resistor is versus or with?

 

[Janiceleong26]

Once you apply KVL over the loop, you'll figure out the current's value and its direction is COUNTER clock wise. Now, if you want to add two series voltages up, you simply need to add the values of these voltages up, BUT you MUST consider the polarities.
In your case,
Cell A:
Since the current direction is counter clock wise, then it will pass Cell A from the right to the left. It will meet the resistance first (with positive sign)and then it will meet the negative terminal of the voltage source (negative sign).
(0.24*2.3 - 4.4 ) = - 3.848 V , the negative sign means the current will first meet the negative terminal of the new voltage.

Cell B:
The current will meet first the positive terminal of the voltage source (positive sign) and then it will find the resistance (positive sign too). As a result, we do the following
(0.24*1.8 + 2.1) = 2.532 V, similarly, the positive sign means the current will initially meet the positive terminal of the new voltage.

I hope this helps a bit.

Best regrads,
G.P.

For Cell A:
I thought Counter clockwise would be from left to right?
For cell B:
Which is the new voltage you are referring to?

 

[GoodPost]

Hhhh ...
I dunno, in my room the clock moves from left to right ><
...
New voltage = voltage source + voltage across the resistor inside the cell.

G.P.

 

[Janiceleong26]

Hhhh ...
I dunno, in my room the clock moves from left to right ><
...
New voltage = voltage source + voltage across the resistor inside the cell.

G.P.

Huh? I thought all clocks moves from right to left?? I.e. Clockwise??
Ok thanks so much

 

[William White]

if this is confusing you do it the easy way

take out one of the batteries (leave its resistance in )

you will have a 4.4V source feeding a 9.6 ohm resistance. Work out the current A

swap the battery, you have 2.1 V source feeding a 9.6 ohm resistance. Work out the current B

total current = A - B (they subtract rather than add because the batteries oppose)now the PDs you want are just ohms law. you know current i; you know resistance R

 

[GoodPost]

Huh? I thought all clocks moves from right to left?? I.e. Clockwise??

^^

Ok thanks so much

You're welcome, all the best wishes.

G.P.

 

[Janiceleong26]

if this is confusing you do it the easy way

take out one of the batteries (leave its resistance in )

you will have a 4.4V source feeding a 9.6 ohm resistance. Work out the current A

swap the battery, you have 2.1 V source feeding a 9.6 ohm resistance. Work out the current B

total current = A - B (they subtract rather than add because the batteries oppose)now the PDs you want are just ohms law. you know current i; you know resistance R

Thanks! That was clear! :)