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How to relate Kirchhoff's law to this question? watch

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Size:  52.4 KB my current working so far: 0.50ohm * 4.0 I = 2 V
    12 - 2 = 10V. The voltage dissipated by all the resistors in that section of the circuit. But am stump since I don't know the resistance value of the others. And I don't really understand the idea about resistance S. Thank you. Thanks!

    Ans: 4.0 Ohms
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    You know that the ratio of resistance of R to other resistor is 1:3, so you can find out the resistance of each resistor in terms of R, and with the help of the fact that the current into a junction equals the current out and the voltage across parallel components is the same, can you work out what R will be?
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    (Original post by zattyzatzat)
    Name:  Temporary.png
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Size:  52.4 KB my current working so far: 0.50ohm * 4.0 I = 2 V
    12 - 2 = 10V. The voltage dissipated by all the resistors in that section of the circuit. But am stump since I don't know the resistance value of the others. And I don't really understand the idea about resistance S. Thank you. Thanks!

    Ans: 4.0 Ohms
    You know that 1A goes through the first resistance of R. How much current goes through the second? What voltage drop, in terms of R, is across each? They must sum to 10V.
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    (Original post by RogerOxon)
    You know that 1A goes through the first resistance of R. How much current goes through the second? What voltage drop, in terms of R, is across each? They must sum to 10V.
    I am a bit confused by this. Would you include the voltage across S as well or not? If so how would you work out that drop and if not, why not?
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    (Original post by RogerOxon)
    You know that 1A goes through the first resistance of R. How much current goes through the second? What voltage drop, in terms of R, is across each? They must sum to 10V.
    I understand how to get the answer now thanks! But as Radioactivedecay said, do you not consider the voltage dissipated across resistor S?
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    (Original post by zattyzatzat)
    I understand how to get the answer now thanks! But as Radioactivedecay said, do you not consider the voltage dissipated across resistor S?
    You can consider the voltage drops across components on any path through the circuit - they must all sum to 10V.
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    (Original post by RogerOxon)
    You can consider the voltage drops across components on any path through the circuit - they must all sum to 10V.
    How though?
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    (Original post by zattyzatzat)
    I understand how to get the answer now thanks! But as Radioactivedecay said, do you not consider the voltage dissipated across resistor S?
    Would you mind telling me where you got the question from? Id like to try questions like these, thanks alot.
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    (Original post by Radioactivedecay)
    Would you mind telling me where you got the question from? Id like to try questions like these, thanks alot.
    CIE A-Levels Physics 2017/Oct/Nov/12 Paper
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    there are three variables - you need 3 equations . i have formed 2 equations , cant seem to form the 3rd
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    (Original post by Radioactivedecay)
    You know that the ratio of resistance of R to other resistor is 1:3
    No, you don't. You know that the voltage at the top of S is different from that at the bottom, as there is a current through it, so the division of current 3:1 implies nothing about the ratio of the resistances.
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    (Original post by RogerOxon)
    No, you don't. You know that the voltage at the top of S is different from that at the bottom, as there is a current through it, so the division of current 3:1 implies nothing about the ratio of the resistances.
    But the way current splits, you know that R has a resistance three times that of the one above it, hence ratio is 1:3.
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    Electricity is very poorly taught IMO. The concepts are actually quite simple, but often not well-explained.

    (Original post by Radioactivedecay)
    I am a bit confused by this. Would you include the voltage across S as well or not? If so how would you work out that drop and if not, why not?
    (Original post by zattyzatzat)
    I understand how to get the answer now thanks! But as Radioactivedecay said, do you not consider the voltage dissipated across resistor S?
    The question tells you to use Kirchhoff's laws. If we look at the 4A flowing into the resistor complex, we see that 3A goes through the first resistor at the top, so 1A goes to the first R. At the point where the R resistors and the S meet, we see that an extra 0.5A is injected by S, so 1.5A flows through the second R. Therefore, our battery voltage (10V) must equal 1R+1.5R=2.5R, so R=4 Ohm.

    Voltage is also known as Electro-Motive Force. It's the force that drives the current through a resistance, just like water pressure drives water flow through your plumbing. Just as the pressure in plumbing, the voltage at a point is the same for all branches that connect to it. The battery voltage of 10V MUST be dropped through all paths through the circuit. So, the drop across both the top resistors must sum to 10V, as must the drop across the two bottom ones, or even the first top resistor, S, and the last bottom resistor.
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    (Original post by RogerOxon)
    Electricity is very poorly taught IMO. The concepts are actually quite simple, but often not well-explained.





    The question tells you to use Kirchhoff's laws. If we look at the 4A flowing into the resistor complex, we see that 3A goes through the first resistor at the top, so 1A goes to the first R. At the point where the R resistors and the S meet, we see that an extra 0.5A is injected by S, so 1.5A flows through the second R. Therefore, our battery voltage (10V) must equal 1R+1.5R=2.5R, so R=4 Ohm.

    Voltage is also known as Electro-Motive Force. It's the force that drives the current through a resistance, just like water pressure drives water flow through your plumbing. Just as the pressure in plumbing, the voltage at a point is the same for all branches that connect to it. The battery voltage of 10V MUST be dropped through all paths through the circuit. So, the drop across both the top resistors must sum to 10V, as must the drop across the two bottom ones, or even the first top resistor, S, and the last bottom resistor.
    if we consider first resistor R then R = 10 OHMS??? V=IR ?
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    (Original post by kiratalreja)
    if we consider first resistor R then R = 10 OHMS??? V=IR ?
    No. The voltage on its left is 10V, the voltage on its right is 6V, so the voltage across it - it's the difference that drives the current - is 4V.
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    (Original post by Radioactivedecay)
    You know that the ratio of resistance of R to other resistor is 1:3, so you can find out the resistance of each resistor in terms of R, and with the help of the fact that the current into a junction equals the current out and the voltage across parallel components is the same, can you work out what R will be?
    (Original post by RogerOxon)
    No, you don't. You know that the voltage at the top of S is different from that at the bottom, as there is a current through it, so the division of current 3:1 implies nothing about the ratio of the resistances.
    (Original post by Radioactivedecay)
    But the way current splits, you know that R has a resistance three times that of the one above it, hence ratio is 1:3.
    I agree with what RogerOxon had replied. Your (I mean Radioactivedecay) argument is based on a parallel circuit setup as shown below.

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    The p.d. across the two resistors is the same. If the ratio of currents I1 : I2 is 3 : 1, then your argument is correct in saying that the resistance of R2 is 3 times greater than the resistance of R1.

    However, the situation in the given circuit is different. See the circuit below.
    The p.d. across KN is different from p.d. across LM. So your argument is not valid.
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    To solve the problem, ensure that you are familiar with Kirchhoff Voltage rule or loop rule and sometimes it is called law instead of rule. Here are 2 videos on the Kirchhoff Voltage rule:
    https://www.youtube.com/watch?v=DM5AWlGT9fE
    https://www.youtube.com/watch?v=ZDoyIghUI44

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    If we traverse the loop ELMYFE, we can write the KVL as following,
    −(1.0 A)R – (1.5 A)R – (4.0A)(0.50 Ω) + 12.0 V = 0
    Solve R as 4.0 Ω
 
 
 
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