Help with vcaa 2012 exam 1 question

[95atar]

had a prob
lem with question 3a and b on the section 2 electricity , annoying as hell.
Very much appreciated

[brenden]

You should provide a link to the exam to make it easier for whoever decided to answer.

[Mr. Study]

I'm going to help you out. However, next time, try and word your post a little better, it's a little ambiguous. I couldn't tell what you were specifically having trouble with and hence, I had to 'walk through' everything, and some bits you may already know.

Question 3 A
So we have the current 2.5 mA. However, this isn't in SI units, so convert to Amps. (2.5 x 10^-3 A).

Now, think about what we have. In the diagram above, we are given a 10 V Battery, a 3000 Ohm Resistor, an LDR (Which a graph is given for it as well) and the total current in the circuit.

We want to know the intensity of light, the lux, falling on the LDR. If we were to look at the graph, given for the LDR, we can see that it is a Resistance (Ohms) Vs Lux graph.

So, if we can find the resistance of the LDR, we can find the lux, or the light intensity, on the LDR.

The total current of the circuit is 2.5 x 10^-3 A. Hence, we can use Ohm's Law.



We know Voltage and we know Current. However, we do not know Resistance. Think about this, as the circuit is in a series, the value of R, or Resistance, will be equal to the sum of the resistance of the resistor and the LDR. Let us call the resistance of the LDR .

Therefore:







Therefore, the resistance of our LDR is 1000 Ohms.

Now, looking at the graph provided, we can see that a resistance of 1000 Ohms (Or 1 kOhm) correlates to a Lux of 10.

Question 3 B
So we are essentially going to design our own circuit. Now, we can only use the LDR, 10 V Battery, the Switching Circuit and a Resistor. Note: We can choose the value of our resistor.

Now, we want the switching circuit to turn on when the LDR has a light intensity of 2.5 or less lux. (2.5 or less lux correlates to a resistance of 3 Ohms, to Infinity)

And we should note that the switching circuit only turns on when there is a voltage, across the input, greater than 6.0 V.

A lot to think about, but .. tbh, I would do this section in pencil and experiment. Firstly, chuck everything into a series shown below but do not add the switching circuit yet, it will just complicate things for now.



Now, let us call the define the resistor's ... resistance (Get it? ) as and the LDR's resistance as .

They are our two unknowns. Now, this is definitely a little tricky but think about this, the resistor will have a set value, as it's resistance is not dependent on the intensity of light. However, the LDR's resistance is dependent on light and, from the graph, it will exponentially increase as Lux decreases, which is what we want. If we were to connect the switching circuit to the resistor, and the light intensity were to decrease, which would cause the resistance of the LDR to increase exponentially, then the switching circuit will not have a voltage greater than 6.0V.

Why?

Remember, this is a series circuit. So the total voltage is equal to the sum of the voltage 'potential', I can't explicitly remember what you label the 'voltage' of each 'item' in the circuit, of the LDR and resistor. So, the higher the resistance, the higher the amount of voltage the item 'takes up'. (Sorry, I really cannot remember the specific Physic's 'terms').

So it would make sense to connect the series circuit to the LDR, as lux decreases, the resistance increases exponentially and hence, taking a large 'chunk' of the total voltage.

We should modify our diagram accordingly.



Now, we do not know the resistance of the resistor or the LDR. However, we do know that we are 'working' with a lux of 2.5 or less. Hence, the absolute minimum value of our LDR's resistance will be 3000 Ohms (or 3kOhms). Hence, we should work out the resistance of the resistor in such a way as to 'give' the LDR a voltage drop (I can't remember the correct term...) of 6 Volts.

To do this, we can work it out by using ... percentages.

Total Voltage is 10.
LDR 'takes up' 60% of the total voltage.
Hence, the resistor will take up 40% of the voltage.

To work out the resistance of the resistor, we can do this:







Hence, the resistance of our resistor will be 2000 Ohms. Therefore our diagram will look like:



Just a note, from this set up our resistor will have a set value of 2000 Ohms. The absolute minimum value of our LDR will be 3000. However, the question states that we want the series circuit to still light up when the lux is less than 2.5. As lux decreases, the resistance of the LDR increases and the resistance of the resistor will stay the same. Hence, the voltage drop of the LDR will increase, above the 6 Volts needed, and the voltage drop of the resistor will decrease.

Hope that helps. =)