# A Kirchoff's Law Question

#### rdb1

Joined Feb 6, 2019
54
So I am working on an example question which has three resistors, 1.5ohms, 4 ohms and 12 ohms connected in parallel, across a 10 volt supply, so I have to calculate the current flowing through each resistor and prove Kirchoff Current Law.

So the first step I did was to use OHMS Law as I need to find the current of each.

So R1 V(10) / R(1.5) = 6.6 R2 V(10) / R (4) = 2.5 R3 V(10) / R(12) = 0.12

So then if I add them all together I get 9.93

Is this correct?

#### dl324

Joined Mar 30, 2015
13,264

#### rdb1

Joined Feb 6, 2019
54
Sorry V is 10 so that's the same

R1 is 1.5 and the Current is 6.6 or 20/3
R2 is 4 and the Current is 2.5 or 5/2
R3 is 12 and the Current is 0.83 or 5/6

To find the Current you do Voltage divided by the Resistance using OHMS Law.

#### dl324

Joined Mar 30, 2015
13,264
To find the Current you do Voltage divided by the Resistance using OHMS Law.
Yes.

What is 20/3 + 5/2 + 5/6? Is it 9.93?

BTW, you have a typo in your first post. 10/12 is not 0.12.

• rdb1

#### rdb1

Joined Feb 6, 2019
54
Yes.

What is 20/3 + 5/2 + 5/6? Is it 9.93?

BTW, you have a typo in your first post. 10/12 is not 0.12.
When you add them all together its 10.

#### rdb1

Joined Feb 6, 2019
54
So there is another

Voltage is 144

R1 is 18 Current is 8
R2 is 20 Current is 36/5
R3 is 30 Current is 24/5

But then when I add the Current together I get 20.

#### Jony130

Joined Feb 17, 2009
5,244
Voltage is 144

R1 is 18 Current is 8
R2 is 20 Current is 36/5
R3 is 30 Current is 24/5

But then when I add the Current together I get 20.
And where is the problem then?

#### rdb1

Joined Feb 6, 2019
54
And where is the problem then?
It requires that you calculate the current flowing through each resistor and prove Kirchoff's Current Law. So then how would you prove that?

#### Jony130

Joined Feb 17, 2009
5,244
Are you trying to tell me that the Itot = 20A is a wrong answer?

#### WBahn

Joined Mar 31, 2012
26,398
So I am working on an example question which has three resistors, 1.5ohms, 4 ohms and 12 ohms connected in parallel, across a 10 volt supply, so I have to calculate the current flowing through each resistor and prove Kirchoff Current Law.

So the first step I did was to use OHMS Law as I need to find the current of each.

So R1 V(10) / R(1.5) = 6.6 R2 V(10) / R (4) = 2.5 R3 V(10) / R(12) = 0.12

So then if I add them all together I get 9.93

Is this correct?
As already noted, you've got some arithmetic problems.

But aside form that, how do you plan to use this to "prove" Kirchhoff's Current Law?

If you find the current in each resistor and add them up, you get 10 A. Fine. So what? How does that prove KCL?

If you determine that the total current from the voltage source is 10 A and then, from that, that KCL is satisfied at that node, does that somehow prove KCL?

No, it doesn't. Because how did you determine that the current from the voltage source is 10 A? If you did it by finding the equivalent resistance of all three resistors in parallel, then you can't use that to prove KCL because the formula that you used to find the equivalent resistance was based on the assumption that KCL is valid. Hence the "proof" is circular.

#### rdb1

Joined Feb 6, 2019
54
As already noted, you've got some arithmetic problems.

But aside form that, how do you plan to use this to "prove" Kirchhoff's Current Law?

If you find the current in each resistor and add them up, you get 10 A. Fine. So what? How does that prove KCL?

If you determine that the total current from the voltage source is 10 A and then, from that, that KCL is satisfied at that node, does that somehow prove KCL?

No, it doesn't. Because how did you determine that the current from the voltage source is 10 A? If you did it by finding the equivalent resistance of all three resistors in parallel, then you can't use that to prove KCL because the formula that you used to find the equivalent resistance was based on the assumption that KCL is valid. Hence the "proof" is circular.
The current is stated in the question as 10A from the supply it then gives you the Resistors in Parallel.

#### WBahn

Joined Mar 31, 2012
26,398
The current is stated in the question as 10A from the supply it then gives you the Resistors in Parallel.
Okay... if you accept the 10 A total as a given, then summing up the individual currents and getting 10 A is less objectionable. I still wouldn't call it a "proof" of KCL, but would phrase it as verifying that KCL is satisfied.

• rdb1

#### rdb1

Joined Feb 6, 2019
54
Okay... if you accept the 10 A total as a given, then summing up the individual currents and getting 10 A is less objectionable. I still wouldn't call it a "proof" of KCL, but would phrase it as verifying that KCL is satisfied.
Okay, if you look at my other question post 6. When I try and do the same I get 20. When really I would assume that it should be 144.

#### WBahn

Joined Mar 31, 2012
26,398
Okay, if you look at my other question post 6. When I try and do the same I get 20. When really I would assume that it should be 144.
Here is where sloppiness with units is biting you.

So let's take a look.

So there is another

Voltage is 144

R1 is 18 Current is 8
R2 is 20 Current is 36/5
R3 is 30 Current is 24/5

But then when I add the Current together I get 20.
You say that the voltage is 144.

But it's not. Just as my height is not 72.

The voltage is 144 V.

Units are a critical part of any physical quantity.

Then you say that R1 is 18 and that Current is 8.

Neither of these are correct.

R1 is 18 Ω. Current is 8 A.

Similarly,

R2 is 20 Ω. Current is 36/5 A
R3 is 30 Ω. Current is 24/5 A

When you add the Currents together, you get 20 A.

Do you see where there is a problem trying to compare 20 A to 144 V?

Do you see what it is you should be comparing here?

#### jayanthd

Joined Jul 4, 2015
906

#### rdb1

Joined Feb 6, 2019
54
Here is where sloppiness with units is biting you.

So let's take a look.

You say that the voltage is 144.

But it's not. Just as my height is not 72.

The voltage is 144 V.

Units are a critical part of any physical quantity.

Then you say that R1 is 18 and that Current is 8.

Neither of these are correct.

R1 is 18 Ω. Current is 8 A.

Similarly,

R2 is 20 Ω. Current is 36/5 A
R3 is 30 Ω. Current is 24/5 A

When you add the Currents together, you get 20 A.

Do you see where there is a problem trying to compare 20 A to 144 V?

Do you see what it is you should be comparing here?
I see what you mean.

#### rdb1

Joined Feb 6, 2019
54
Battery had internal resistance of 0.1R in my previous simulation. I removed it. Here are the results.

View attachment 169739

View attachment 169740
Thanks for that I see were I was going wrong. You need to find the total current in this case 20A, then if you look at the amps going into each of the resistors it should add up to the total current.

#### DrewStupid

Joined Nov 28, 2018
64
If the votage drop over each risistor adds up to the supply voltage and the current through each parrallel resistor adds up to total current (IT) it simply proves that your calculations are correct. It does not proof the law !