Solving Circuits: EMF 12V, Internal Resistance 5Ω

[kenshi64]

On the attached image of a circuit I was given a sum in the exam as follows, but I can't make sense of it:

The circuit has an EMF of 12V and an internal resistance of 5ohms.

Q1) Total resistance of the circuit
A) 50(45+5) I don't get why they add 5 since that's the internal resistance!

Q2) Current in the internal resistance
A) I= (12/50) = 0.24. Why is 50 used? the internal resistance is 5 not 50! and EMF is provided too, please explain! :D

Q3)Potential difference between points X and Y
A)across 30ohm voltage drops by 3.60 V (so potential at X is 3.60 V);
across 60ohm voltage drops by 7.20 V (so potential at Y is 7.20 V);
so potential difference between X and Y is (negative) 3.6 V;
---Here I am completely lost as to how they got these values, I'm infinitely grateful for feedback, this sum has been eating away at me for long! :)

 

[MisterX]

You didn't successfully attach the image.

 

[sendthis]

Q1) Total resistance of the circuit
A) 50(45+5) I don't get why they add 5 since that's the internal resistance!

Q2) Current in the internal resistance
A) I= (12/50) = 0.24. Why is 50 used? the internal resistance is 5 not 50! and EMF is provided too, please explain! :D
)

Yeah, with no diagram, I can only guess, but you have two resistors in series. One of them is apparently via a resitor and one is because of the load?

Anyway, since they're in series, the flow of current through one is going to have to be the same as the other.

Since I can't see a circuit, I'm not going to go too deep, but look over the section in your textbook where current going into a node has to equal the current leaving (Kirchhoff's current law (KCL))

hth

 

[sendthis]

Q3)Potential difference between points X and Y
A)across 30ohm voltage drops by 3.60 V (so potential at X is 3.60 V);
across 60ohm voltage drops by 7.20 V (so potential at Y is 7.20 V);
so potential difference between X and Y is (negative) 3.6 V;
---Here I am completely lost as to how they got these values, I'm infinitely grateful for feedback, this sum has been eating away at me for long! :)

Okay, I'll take a stab at Q3 too...

You have
-------R1----------------- R2---------
X----/\/\/\-----(N)------/\/\/\/\/-----Y

R1 = 30ohms
R2 = 60ohms
Rtotal = 90ohms

There is a parallel voltage source of 10.8V?

V(R1) = R1/Rtotal * 10.8 =3.6V
V(R2) = R2/Rtotal * 10.8 =7.2V

The difference between V(R1) and V(R2) = V(R1) - V(R2) = 3.6 - 7.2 = -3.6V
Since the difference can be stated in absolute terms, the difference is actually just 3.6V.

Maybe the way it's stated in the exam, it cares...

 

[kenshi64]

Here's the Image sorry!

 

[sendthis]

Are you not including information? I'm getting a value of 4V. That's a wheatstone bridge.

Is there a resistance between X and Y that's mentioned in the problem that's not mentioned in your post?

 

[MisterX]

If I created that diagram I might have added a resistor for the "internal resistance." If it makes things clearer, draw a new diagram where the supply symbol is replaced by a symbol representing and ideal 12 V source in series a 5 Ohm resistor.

Q1)The total resistance across the ideal voltage source would be 5 + 45 = 50 Ohms. This must be what the author meant by "total resistance."

Q2)The current would be the ideal voltage source's voltage divided by the resistance across it. 12 V/ 50 ohms = 0.24 A.

Q3) Both branches have equal resistance, so the current will divide equally into each branch. Consider the two rightmost resistors. If the voltage to the right of them is V₀.


The voltage at X is V₀ + 30 Ohms * 0.12 A = V₀ + 3.6 V.

The voltage at Y is V₀ + 60 Ohms * 0.12 A = V₀ + 7.2 V

The voltage between X and Y is thus V_X - V_Y = V₀ + 3.6 V - (V₀ + 7.2 V) = -3.6 V

If it helps make things clear, draw an arrow on the diagram above the 30 and 60 Ohm resistors pointing from left to right. This is the direction of the current. Then put a + sign near the tail end of each arrow and a - sign near the head of each arrow. This is to indicate the polarity of the voltage that will be produced across the resistor when the current is flowing in the direction of the arrow.