Kirchhoff's law problem solving - alternative

Question from Physics for the IB Diploma page 228 number 53.

Determine the magnitudes and directions of the currents in the circuit  shown below

Answer:

Kirchhoff's first law: "Total current in to a node/junction = total current leave the node/junction."
I₃ = I₁ + I₂     ................. (1)

Kirchhoff's second law: "Total emf = total potential differences."
Loop kiri
-4 = 2I₁ + 3I₃  ................ (2)

Loop kanan
-6 = 4I₂ + 3I₃   ............... (3)

Substitute (1) to (2)
-4 = 2I₁ + 3I₃ = 2I₁ + 3(I₁ + I₂) = 2I₁ + 3I₁ + 3I₂
-4 = 5I₁ + 3I₂   ............... (4)

Substitute (1) to (3)
-6 = 4I₂ + 3I₃ = 4I₂ + 3(I₁ + I₂) = 4I₂ + 3I₁ +3I₂
-6 = 3I₁ + 7I₂   ............... (5)

Eliminate (4) with (5)
-4 = 5I₁ + 3I₂  (x3)
-6 = 3I₁ + 7I₂  (x5)

-12 = 15I₁ + 9I₂
-30 = 15I₁ + 35I₂   _

---------------------

18 = -26I₂

I₂  = 18 : -26 = -0.69 A (I₂ = 0.69 A direction to the left)

Substitute I₂ value to (3)

-6 = 4(-0.69) + 3I₃

-6 = -2.76 + 3I₃

-6 + 2.76 = 3I₃

-3.24 = 3I₃

I₃ = -3.24 : 3 = -1.08 A (I₃ = 1.08 A direction upwards)

Substitute I₂ and I₃ values to (1)

I₃ = I₁ + I₂

-1.08 = I₁ + (-0.69)

I₁ = -1.08 + 0.69 = -0.39 A (I₁ = 0.39 A direction to the right)

Kirchhoff's law problem solving

Question from Physics for the IB Diploma page 228 number 53.

Determine the magnitudes and directions of the currents in the circuit  shown below

Answer:

Kirchhoff's first law: "Total current in to a node/junction = total current leave the node/junction."
I₃ = I₁ + I₂     ................. (1)

Kirchhoff's second law: "Total emf = total potential differences."
Loop kiri
4 = 3I₃ + 2I₁  ................ (2)

Loop kanan
6 = 4I₂ + 3I₃   ............... (3)

Substitute (1) to (2)
4 = 3I₃ + 2I₁ = 3(I₁ + I₂) + 2I₁ = 3I₁ + 3I₂ + 2I₁
4 = 5I₁ + 3I₂   ............... (4)

Substitute (1) to (3)
6 = 4I₂ + 3I₃ = 4I₂ + 3(I₁ + I₂) = 4I₂ + 3I₁ +3I₂
6 = 3I₁ + 7I₂   ............... (5)

Eliminate (4) with (5)
4 = 5I₁ + 3I₂  (x3)
6 = 3I₁ + 7I₂  (x5)
----------------------

12 = 15I₁ + 9I₂
30 = 15I₁ + 35I₂    _

---------------------

-18 = -26I₂

I₂  = 18 : 26 = 0.69 A (I₂ direction to the left)

Substitute I₂ value to (3)

6 = 4(0.69) + 3I₃

6 = 2.76 + 3I₃

6 - 2.76 = 3I₃

3.24 = 3I₃

I₃ = 3.24 : 3 = 1.08 A (I₃ direction upwards)

Substitute I₂ and I₃ values to (1)

I₃ = I₁ + I₂

1.08 = I₁ + 0.69

I₁ = 1.08 - 0.69 = 0.39 A (I₁ direction to the right)

November 2010 paper 1 question 2 - DP HL

Question:

Two lengths, a and b, are measured to be 51 ± 1 cm and 49 ± 1 cm respectively. In which of the following quantities is the percentage uncertainty the largest?
A. a + b
B. a – b
C. a x b
D. a : b


Answer: B

% uncertainty of a = (1 : 51) x 100% = 2.0 %
% uncertainty of b = (1 : 49) x 100% = 2.0 %

A. a + b = (51 + 49) ± (1 + 1) = 100 ± 2 cm
    % uncertainty of (a + b) = (2 : 100) x 100 % = 2 %

B. a - b = (51 - 49) ± (1 + 1) = 2 ± 2 cm
    % uncertainty of (a - b) = (2 : 2) x 100% = 100 %

C.  & D. 
When we multiply or divide quantities we add their fractional or percentage uncertainties, so:
% uncertainty of (a x b) and (a : b) = 2.0 % + 2.0 % = 4.0 %