Post your calculations, and the results you get, and we'll sober up a bit and see if you are right.
Hi i need help with what equations to use to get the results. I literally have 0 knowledge about this kind of stuff
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Hi i dont have any calculations because i dont know what equations to use or how to go about getting the results
If you’ve not been taught it it’s unreasonable and unrealistic if the course provider to expect you to understand it mate, there are plenty of people on here that could give you the answer....but would that really help you? You need to speak to the tutor mate...there’s no shame in putting ya hand up and saying you don’t get it, after all I assume you or someone else is paying these people to teach you it, when I did my courses it cost thousands and there was no way I was getting fobbed it till they’d taught me it ?
to be fair my tutor as about as helpful as a chocolate fireguard but thats another story. is there an equation or explanation for this that i might be able to follow? you dont have to tell me the answer, just how to do it and then i could try and do the maths on my own.
thanks in advance.
Let's start you off then.
Get your pencil.
Name the voltage where the resistors join (for the sake of argument call it V2).
Name the current in the left hand resistor I1 and in the right hand resistor I2.
Remember Ohm's law. That's it, it really is. - But to take you one step further..
Mr Georg Ohm says that I1 = (12-V2)/1000
and also that I2 = (5-V2)/1500
He also says that (I1+I2) =V2/680
The rest is up to you. Some very basic manipulation, solve for V2, I1 and I2.
Get your pencil.
Name the voltage where the resistors join (for the sake of argument call it V2).
Name the current in the left hand resistor I1 and in the right hand resistor I2.
Remember Ohm's law. That's it, it really is. - But to take you one step further..
Mr Georg Ohm says that I1 = (12-V2)/1000
and also that I2 = (5-V2)/1500
He also says that (I1+I2) =V2/680
The rest is up to you. Some very basic manipulation, solve for V2, I1 and I2.
The question asks for the solution to be found using Kirchoff's laws (presumably to demonstrate an understanding of them.
Kirchoof's current law states that the sum of currents flowing into a circuit node (a point where things connect together) must equal the sum of currents flowing out (i.e. the algebraic sum must be zero).
Kirchoff's voltage law states that the algebraic sum of voltages around a closed loop must be zero.
Take some time to internalise these concepts: they are important to understanding how things behave. After a while they will seem intuitive and you will wonder why Kirchoff got any recognition for stating the totally obvious.
Look at the loop V1-R1-R3. We know V1, and from KVL we know that the sum of the voltages across R1 and R3 must equal V1 in magnitude (taking care with the sign in each case!) The same is true of the other loops: V2-R3-R2 and V1-R1-R2-V2.
From KCL you can say things about the current in each resistor in terms of the currents in the others. As we know the resistance values, we can also write the voltages in terms of the currents through each resistor using Ohm's law. Hence we can write a bunch of simultaneous equations to solve for any of the unknowns.
Just watch the polarities / signs: Decide on a direction and stick to it even if you realise the current might be flowing the other way after all. It doesn't matter provided you are consistent.
Kirchoof's current law states that the sum of currents flowing into a circuit node (a point where things connect together) must equal the sum of currents flowing out (i.e. the algebraic sum must be zero).
Kirchoff's voltage law states that the algebraic sum of voltages around a closed loop must be zero.
Take some time to internalise these concepts: they are important to understanding how things behave. After a while they will seem intuitive and you will wonder why Kirchoff got any recognition for stating the totally obvious.
Look at the loop V1-R1-R3. We know V1, and from KVL we know that the sum of the voltages across R1 and R3 must equal V1 in magnitude (taking care with the sign in each case!) The same is true of the other loops: V2-R3-R2 and V1-R1-R2-V2.
From KCL you can say things about the current in each resistor in terms of the currents in the others. As we know the resistance values, we can also write the voltages in terms of the currents through each resistor using Ohm's law. Hence we can write a bunch of simultaneous equations to solve for any of the unknowns.
Just watch the polarities / signs: Decide on a direction and stick to it even if you realise the current might be flowing the other way after all. It doesn't matter provided you are consistent.
Kirchhoff is really "common sense" applied to Ohm's law. Once you get that it's all the same thing. Tutors may get a bit ---- about it so I understand your point. In this case Kirchhoffs "laws" are used in the way I did it also. (Not being a law at all really- just being a rather obvious observation resulting from Ohm) :
Current into a node = currents out. Hence currents I1 and I2 are summed in the middle resistor which takes I1+I2 (Kirchhoffs first law).
Voltage round a loop equals zero. This is implicit in the first and second and third statements. Insofar as I said:
I1=(12-V2)/1000 . The voltage across the battery is 12V which equals the 12 Volts across the remainder of the loop of 1000 and 680 ohm resistors. Ie 12-12 = 0 (Kirchhoff second law)
Likewise the second. The voltage across the battery is 5V which equals the 5 volts across the 1500 and 680 ohm resistors. Is 5-5 = 0 (Kirchhoff)
Another way of doing the same as Lucien points out might be to set it out more explicitly as your tutor may define as “Kirchhoff” by setting out two clockwise circulating currents (one will be negative), but at the end of the day it’s all Ohms law!
Have fun
Bobby2017: Well done for having a go. You know how to solve a problem using simultaneous equations. Alas, the simultaneous equations you wrote down are incorrect. You made errors because you did not define the currents i1 and i2 and their directions nor the voltage drops across the resistors and their polarity. This meant that when you added up the voltage drops in a loop to equal the electromotive force in the loop you had some signs wrong in the simultaneous equations.
An important lesson then is to define your currents and voltage drops on a diagram as I did top left in the attachment.
Normally in a dc circuit one would use conventional current flow ie from positive to negative. So current 'leaves' the positive terminal of the battery. The polarity of the voltages across a resistor when a current flows through them as I have defined them is as shown by the arrows in my attached diagram where the arrow tip is +ve with respect to its tail - note well that for the currents I have defined the voltage drop polarity (arrow direction) is in the opposite direction.
Applying Kirchoff's voltage law = Sum of emfs in a loop equals sum of voltage drops - for the left hand loop - it would have been helpful if I had remembered to put arrows beside each battery to indicate the direction of the emfs of the batteries. For the 12V battery it would point upwards and for the 5V battery it would point downwards:
12 = 1000i1 + 680(i1-i2)
For the right hand loop:
5 = 1500i2 - 680(i1-i2).......Note well the minus sign before the 680 Ohm resistor because the polarity of the voltage drop I defined across this resistor is in the opposite direction to the voltage drop across the 1500 Ohm resistor.
After deriving the simultaneous equations I then use the matrix algebra method to do the mathematics to arrive at i1 = 9.2375mA and i2 = 5.175mA. Because these are positive values these currents are indeed in the directions I defined in the diagram. A negative sign would reverse the direction of current flow.
One then finds the voltage drop across the 680 Ohm using using Kirchoff's current law. The current through the 680 Ohm resistor is i1-i2, so one can then use Ohm's Law to calculate the voltage drop. And since i1 is greater than i2, (i1-i2) is a positive number which means that the voltage drop polarity across the 680 is indeed in the direction I initially defined. If (i1-i2) was negative this would indicate the voltage drop polarity is opposite to what I defined.
As clear as mud? I hope not.
An important lesson then is to define your currents and voltage drops on a diagram as I did top left in the attachment.
Normally in a dc circuit one would use conventional current flow ie from positive to negative. So current 'leaves' the positive terminal of the battery. The polarity of the voltages across a resistor when a current flows through them as I have defined them is as shown by the arrows in my attached diagram where the arrow tip is +ve with respect to its tail - note well that for the currents I have defined the voltage drop polarity (arrow direction) is in the opposite direction.
Applying Kirchoff's voltage law = Sum of emfs in a loop equals sum of voltage drops - for the left hand loop - it would have been helpful if I had remembered to put arrows beside each battery to indicate the direction of the emfs of the batteries. For the 12V battery it would point upwards and for the 5V battery it would point downwards:
12 = 1000i1 + 680(i1-i2)
For the right hand loop:
5 = 1500i2 - 680(i1-i2).......Note well the minus sign before the 680 Ohm resistor because the polarity of the voltage drop I defined across this resistor is in the opposite direction to the voltage drop across the 1500 Ohm resistor.
After deriving the simultaneous equations I then use the matrix algebra method to do the mathematics to arrive at i1 = 9.2375mA and i2 = 5.175mA. Because these are positive values these currents are indeed in the directions I defined in the diagram. A negative sign would reverse the direction of current flow.
One then finds the voltage drop across the 680 Ohm using using Kirchoff's current law. The current through the 680 Ohm resistor is i1-i2, so one can then use Ohm's Law to calculate the voltage drop. And since i1 is greater than i2, (i1-i2) is a positive number which means that the voltage drop polarity across the 680 is indeed in the direction I initially defined. If (i1-i2) was negative this would indicate the voltage drop polarity is opposite to what I defined.
As clear as mud? I hope not.
Attachments
Last edited:
Thankyou so much for your help ? you are a legend !!Bobby2017: Well done for having a go. You know how to solve a problem using simultaneous equations. Alas, the simultaneous equations you wrote down are incorrect. You made errors because you did not define the currents i1 and i2 and their directions nor the voltage drops across the resistors and their polarity. This meant that when you added up the voltage drops in a loop to equal the electromotive force in the loop you had some signs wrong in the simultaneous equations.
An important lesson then is to define your currents and voltage drops on a diagram as I did top left in the attachment.
Normally in a dc circuit one would use conventional current flow ie from positive to negative. So current 'leaves' the positive terminal of the battery. The polarity of the voltages across a resistor when a current flows through them as I have defined them is as shown by the arrows in my attached diagram where the arrow tip is +ve with respect to its tail - note well that for the currents I have defined the voltage drop polarity (arrow direction) is in the opposite direction.
Applying Kirchoff's voltage law = Sum of emfs in a loop equals sum of voltage drops - for the left hand loop - it would have been helpful if I had remembered to put arrows beside each battery to indicate the direction of the emfs of the batteries. For the 12V battery it would point upwards and for the 5V battery it would point downwards:
12 = 1000i1 + 680(i1-i2)
For the right hand loop:
5 = 1500i2 - 680(i1-i2).......Note well the minus sign before the 680 Ohm resistor because the polarity of the voltage drop I defined across this resistor is in the opposite direction to the voltage drop across the 1500 Ohm resistor.
After deriving the simultaneous equations I then use the matrix algebra method to do the mathematics to arrive at i1 = 9.2375mA and i2 = 5.175mA. Because these are positive values these currents are indeed in the directions I defined in the diagram. A negative sign would reverse the direction of current flow.
One then finds the voltage drop across the 680 Ohm using using Kirchoff's current law. The current through the 680 Ohm resistor is i1-i2, so one can then use Ohm's Law to calculate the voltage drop. And since i1 is greater than i2, (i1-i2) is a positive number which means that the voltage drop polarity across the 680 is indeed in the direction I initially defined. If (i1-i2) was negative this would indicate the voltage drop polarity is opposite to what I defined.
As clear as mud? I hope not.
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