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Preliminary Electricity: Potential in a Circuit (1 Viewer)

RivalryofTroll

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What is the potential at X in the circuit below:

Potential Electricity.png

How to do this? (I need full solution and pointers on how to do these types of questions)
 

bobmcbob365

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If you already know what potential and potential difference is, you can skip the next 2 paragraphs.

So basically, the power supply adds potential energy to the current of the circuit, going anti-clockwise in this case. The total amount of energy added per unit charge is obviously the voltage of the power supply (the battery) (measured in Joules per coloumb or Volt).

As the current goes through the circuit, some energy is used up when it crosses a resistor. This is because, some of the energy is converted into heat (or light if it is a light globe etc.). So therefore, the amount of energy left after going through a resistor is less than what it was before it passed through. This drop in electric potential energy is called the voltage drop. (This is what you actually see when you connect a voltmeter in parallel with a certain resistor - merely measuring the difference in potential energy between two points, also called potential difference)

So in questions like this, you almost always try to find V, I and R (V being the electromotive force (or the potential difference across the power supply), I being the electric current and R being the resistance), so that you can use the famous Ohm's law: V = IR.

In this question, you are given the electromotive force and resistance of the circuit, i.e. V = 24, R1 = 4, R2 = 2.

So first you measure out the total resistance in the circuit. In this case, the resistors are connected in series, so total resistance = 2 + 4 = 6 Ohm.

Now, you can use the formula I = V/R to work out the current in the circuit:

I = 24 / 6
= 4 Amps

Now that you know the current of the circuit. You can then use V = IR again to find the voltage drop for a single resistor (the 4 Ohm resistor). (Since the resistors are in series, the current is the same as the current measured previously)

V = 4 * 4
= 16

So the potential difference or the voltage drop across the 4 Ohmic resistor is 16 volts. This also means that 16 Joules of potential energy is lost and converted into mainly heat energy. Since the power supply supplies 24 V or in other words, supplies 24 Joules of potential energy, the amount of potential energy after passing through the resistor (the point at X) is 24 minus 16 which equals to 8 Volts.

So basically, potential, in these instances, is the amount of energy that is left after passing through a resistor.
Initially at 24 V at the positive end of power supply, potential gradually decreases as you go along the conventional current of the circuit, and reaches 0 V at the negative end of the power supply.

So if you ever encounter questions like this, the main things to do are:
1. Find V, I and R
2. Use V = IR to find the potential difference across a resistor.
3. Minus this from the voltage of the power supply.
4. If there are more than one resistor between the positive end of the cell/ battery to whatever point you are asked, simply continue subtracting the potential difference across each resistor until you reach the point you are asked.
5. This will get you the potential at any point in a circuit.


If anything confuses you or if you need clarity, feel free to ask.


edit: i added a picture which might help you understand potential.
 

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such_such

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Another way:

1. Find total resistance
= 2 + 4
= 6 Ohms

2. Find total current
V = IR
24 = 6I
I = 4 A

3. Find voltage at 2 Ohms
V = IR
V = 4 x 2
V = 8 V

You go from higher potential (positive) to lower potential (negative), so in your circuit, you go from left to right.

You only go through the 2 Ohm resistor

At the 2 Ohm resistor, you lose 8V (calculated above) [you 'lose' voltage when you go through a resistor]

Since you started off with 24V, now you lose 8V when you pass through the 2 Ohm

4. Calculate: voltage supply - voltage drop
= 24 - 8
= 16 V
 
Last edited:

Living Moment

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If you already know what potential and potential difference is, you can skip the next 2 paragraphs.

So basically, the power supply adds potential energy to the current of the circuit, going anti-clockwise in this case. The total amount of energy added per unit charge is obviously the voltage of the power supply (the battery) (measured in Joules per coloumb or Volt).

As the current goes through the circuit, some energy is used up when it crosses a resistor. This is because, some of the energy is converted into heat (or light if it is a light globe etc.). So therefore, the amount of energy left after going through a resistor is less than what it was before it passed through. This drop in electric potential energy is called the voltage drop. (This is what you actually see when you connect a voltmeter in parallel with a certain resistor - merely measuring the difference in potential energy between two points, also called potential difference)

So in questions like this, you almost always try to find V, I and R (V being the electromotive force (or the potential difference across the power supply), I being the electric current and R being the resistance), so that you can use the famous Ohm's law: V = IR.

In this question, you are given the electromotive force and resistance of the circuit, i.e. V = 24, R1 = 4, R2 = 2.

So first you measure out the total resistance in the circuit. In this case, the resistors are connected in series, so total resistance = 2 + 4 = 6 Ohm.

Now, you can use the formula I = V/R to work out the current in the circuit:

I = 24 / 6
= 4 Amps

Now that you know the current of the circuit. You can then use V = IR again to find the voltage drop for a single resistor (the 4 Ohm resistor). (Since the resistors are in series, the current is the same as the current measured previously)

V = 4 * 4
= 16

So the potential difference or the voltage drop across the 4 Ohmic resistor is 16 volts. This also means that 16 Joules of potential energy is lost and converted into mainly heat energy. Since the power supply supplies 24 V or in other words, supplies 24 Joules of potential energy, the amount of potential energy after passing through the resistor (the point at X) is 24 minus 16 which equals to 8 Volts.

So basically, potential, in these instances, is the amount of energy that is left after passing through a resistor.
Initially at 24 V at the positive end of power supply, potential gradually decreases as you go along the conventional current of the circuit, and reaches 0 V at the negative end of the power supply.

So if you ever encounter questions like this, the main things to do are:
1. Find V, I and R
2. Use V = IR to find the potential difference across a resistor.
3. Minus this from the voltage of the power supply.
4. If there are more than one resistor between the positive end of the cell/ battery to whatever point you are asked, simply continue subtracting the potential difference across each resistor until you reach the point you are asked.
5. This will get you the potential at any point in a circuit.


If anything confuses you or if you need clarity, feel free to ask.


edit: i added a picture which might help you understand potential.
Chill out guys this isn't going to be in the the physics HSC paper
 

Living Moment

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What's it to you?

You seem to be doing HSC Physics, so maybe you could help OP out.
What is it with me? guess I just trying to help out a brother :D

The truth is Prelim doesn’t matter one bit in the HSC course which is a shame, Honestly if I could of went back in time I would of study the HSC course in prelim

IF you have any HSC Q. I'll be happy to help :)
 

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