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

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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

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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#2

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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#3

(Original post by

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

**zattyzatzat**)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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#4

(Original post by

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.

**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.

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**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.

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#6

(Original post by

I understand how to get the answer now thanks! But as Radioactivedecay said, do you not consider the voltage dissipated across resistor S?

**zattyzatzat**)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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#7

(Original post by

You can consider the voltage drops across components on any path through the circuit - they must all sum to 10V.

**RogerOxon**)You can consider the voltage drops across components on any path through the circuit - they must all sum to 10V.

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#8

**zattyzatzat**)

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

Would you mind telling me where you got the question from? Id like to try questions like these, thanks alot.

**Radioactivedecay**)Would you mind telling me where you got the question from? Id like to try questions like these, thanks alot.

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#10

there are three variables - you need 3 equations . i have formed 2 equations , cant seem to form the 3rd

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#11

(Original post by

You know that the ratio of resistance of R to other resistor is 1:3

**Radioactivedecay**)You know that the ratio of resistance of R to other resistor is 1:3

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#12

(Original post by

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.

**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.

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#13

Electricity is very poorly taught IMO. The concepts are actually quite simple, but often not well-explained.

(Original post by

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.

(Original post by

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?

**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?

**zattyzatzat**)

I understand how to get the answer now thanks! But as Radioactivedecay said, do you not consider the voltage dissipated across resistor S?

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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#14

(Original post by

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.

**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.

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#15

(Original post by

if we consider first resistor R then R = 10 OHMS??? V=IR ?

**kiratalreja**)if we consider first resistor R then R = 10 OHMS??? V=IR ?

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#16

(Original post by

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?

**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?

**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

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.

**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.

The p.d. across the two resistors is the same. If the ratio of currents

*I*

_{1}:

*I*

_{2}is 3 : 1, then your argument is correct in saying that the resistance of

*R*

_{2}is 3 times greater than the resistance of

*R*

_{1}.

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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#17

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

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

https://www.youtube.com/watch?v=DM5AWlGT9fE

https://www.youtube.com/watch?v=ZDoyIghUI44

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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