For anyone not understanding Q2b, you can also write it as: 9-epsilon-9.8(I2)-1.2(I1)=0 9-epsilon-9.8(0.133)-1.2(0.333)=0 -epsilon=-7.297 epsilon (EMF) = 7.3v
Sure Kevin! Before the circuit changes the voltage across both the voltage source and capacitor is 6V, their voltages balance and no current flows. When the capacitance is changed the capacitor voltage reduces to 3.05V, so that now the voltage is unbalanced by 6V-3.05V. To determine the current flow this causes we can use I=V/R where the resistance in the circuit is still 12Ω, but the circuit voltage is now 6V-3.05V as mentioned. This gives us a current flow of 0.246A.
Hi Eli, very good question. The circuit does not contain a capacitor so XC=0. This means that the XC term in √(R^2 + (XL-XC)^2) disappears and we're left with √(R^2 + XL^2).
@@MrWhibleyPhysics Thank you, I had another quick question if you would be able to help. When calculating points on a graph of charging/discharging, when do you multiply by 0.63 and when do you multiply by 0.37 in order to calculate the point? Thank you.
@@elic3286 The time constant is always the time for a 63% change. If the value is increasing it will increase to 63% of its maximum value. If the value is decreasing it will decrease by 63% of its maximum value, at which point it will be 37% (100% - 63%) of its maximum value.
For question 2b when it said calculate the maximum magnetic flux in each of the three coils of wire, What I did was I divided the result of (0.0413 x (1.60m x 0.600m)) by 3. Why I did this was when the question said each I thought I was only meant to find 1 magnetic flux of each coil. Where did I possibly went wrong?
@@MrWhibleyPhysics Thank you that was helpful. For an example if a similar question like that says in 3 of the 3 coils will I have to divide by 3. For these types of questions do you divide by 3?
@@MrWhibleyPhysics thank you! one last question (for now haha), i dont understand why the V used in the last question is the difference between 6 and 3.05? why is it not just 6 instead- is it not a max Voltage for max Current type of situation?
@@aimisofea7964 Great question! I think I could have explained this better. Consider that the circuit current is the same as the resistor current since they are both in series. So we can approach this as finding the resistor current, using I=V/R with V as the resistor voltage and R as the resistor's resistance. We know that the resistance R is 12.0Ω. We know that the resistor voltage + capacitor voltage = source voltage. Since the capacitor voltage is 3.05V and the source voltage is 6V, this becomes... resistor voltage + 3.05V = 6V or resistor voltage = 6V-3.05V Hope that helps!
@@nhs_gtk8142 0.4V is the voltage drop across the 1.2ohm resistor. In our voltage equation, subtracting the 1.2ohm resistor means subtracting its 0.4V voltage drop.
![กี่หมื่นฟ้า | Your Sky Series EP.1 [1/4]](http://i.ytimg.com/vi/vLGjxFL6U_I/mqdefault.jpg)
real respect for the teacher... really helpful...
Look had exam today, perfect resource for last minute marking before exam. Love it!
That's awesome to hear Seth, hope it went well!
For anyone not understanding Q2b, you can also write it as:
9-epsilon-9.8(I2)-1.2(I1)=0
9-epsilon-9.8(0.133)-1.2(0.333)=0
-epsilon=-7.297
epsilon (EMF) = 7.3v
Goat!!!
Hey can you explain your calculation on the 3d a bit more? It doesn't really make sense to me. Thanks
Sure Kevin! Before the circuit changes the voltage across both the voltage source and capacitor is 6V, their voltages balance and no current flows. When the capacitance is changed the capacitor voltage reduces to 3.05V, so that now the voltage is unbalanced by 6V-3.05V. To determine the current flow this causes we can use I=V/R where the resistance in the circuit is still 12Ω, but the circuit voltage is now 6V-3.05V as mentioned. This gives us a current flow of 0.246A.
Hi, for question 2d, how come you use the equation for impedance as Z=square root (R^2 + XL^2), instead of square root (R^2 + (XL+XC)^2)? Thank you.
Hi Eli, very good question. The circuit does not contain a capacitor so XC=0. This means that the XC term in √(R^2 + (XL-XC)^2) disappears and we're left with √(R^2 + XL^2).
@@MrWhibleyPhysics Thank you, I had another quick question if you would be able to help. When calculating points on a graph of charging/discharging, when do you multiply by 0.63 and when do you multiply by 0.37 in order to calculate the point? Thank you.
@@elic3286 The time constant is always the time for a 63% change. If the value is increasing it will increase to 63% of its maximum value. If the value is decreasing it will decrease by 63% of its maximum value, at which point it will be 37% (100% - 63%) of its maximum value.
Can u go over the 2018 ncea lvl 3 electricity paper
Hi, why do we use the left-hand loop for quesion 1b, Kirchoff's Volage Law, instead of the right-hand loop? Thanks
Hi Christine. Both should work, I chose the left loop because it required less additional calculating.
Hi Mr Whibley, I was also wondering if you are able to make a walkthrough on 2018 electrical systems? (Level 3)@@MrWhibleyPhysics
For question 2b when it said calculate the maximum magnetic flux in each of the three coils of wire, What I did was I divided the result of (0.0413 x (1.60m x 0.600m)) by 3. Why I did this was when the question said each I thought I was only meant to find 1 magnetic flux of each coil. Where did I possibly went wrong?
Hi AnanomousRex, because we're using the area of 1 loop already (1.60m x 0.600m) there is no need to divide it by 3.
@@MrWhibleyPhysics Thank you that was helpful. For an example if a similar question like that says in 3 of the 3 coils will I have to divide by 3. For these types of questions do you divide by 3?
Amazing video this is really going to help
Thank you atarangi5477, glad to hear it!
for 2c, why does decreasing Xl cause a decrease in Z?
Hi Aimi. Z^2=X^2 + R^2, so a reduction in X will result in a reduction in Z.
@@MrWhibleyPhysics thank you! one last question (for now haha), i dont understand why the V used in the last question is the difference between 6 and 3.05? why is it not just 6 instead- is it not a max Voltage for max Current type of situation?
@@aimisofea7964 Great question! I think I could have explained this better.
Consider that the circuit current is the same as the resistor current since they are both in series. So we can approach this as finding the resistor current, using I=V/R with V as the resistor voltage and R as the resistor's resistance.
We know that the resistance R is 12.0Ω.
We know that the resistor voltage + capacitor voltage = source voltage. Since the capacitor voltage is 3.05V and the source voltage is 6V, this becomes...
resistor voltage + 3.05V = 6V
or
resistor voltage = 6V-3.05V
Hope that helps!
@@MrWhibleyPhysics ahhh, i see! thank you so much ^◡^
For 1b why didnt we subtract the 1.2 ohms
Not sure if I've correctly understood your question. In both cases we did subtract the 0.4V of the 1.2ohm resistor.
@@MrWhibleyPhysics but why didnt we subtract the 1.2 ohm resistor and jus the voltage drop? The one near the 9v
@@nhs_gtk8142 0.4V is the voltage drop across the 1.2ohm resistor. In our voltage equation, subtracting the 1.2ohm resistor means subtracting its 0.4V voltage drop.
@@MrWhibleyPhysics ohhh right tysm!!!
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