Engineering sceince - Essay Example

Now according to Kirchhoff’s current law (KCL), the algebraic sum of all the currents
entering and leaving any node in a circuit is zero (Zahavi, and Edminister 24). No let us neglect
for time being both the voltage sources and considering the flow of current as shown in fig.3.1.2,
then the current at node A is given as;
I1- I2 – I3 = 0 …. (1)

Also according to Kirchhoff’s voltage law (KVL), the algebraic sum of the voltage rises, and voltage drops around a loop must be zero (Zahavi, and Edminister 24). Again considering the fig. 3.1.2 for loops L1 and L2 and applying the KVL while looking in the clockwise direction, we have;
For L1, - 24I1 - 13I2 - 85=0 … (2)
24I1 + 13I2 + 85=0 … (3)
And for loop L2,
-85+13I2 + 10I3 + 110=0
Or, 13I2 + 10I3 + 25=0 … (3)
Now from equation (1),
I1= I2 + I3 ... (4)
Substituting this value in equation (2),
24(I2 + I3) +13I2 - 85=0
Or, 37I2 + 24I3 + 85=0 … (5)
From equations (3) and (5), solving for I2,
I2 = 4.31 A,
Substitution I2, value in equation (5);
I3 = 0.46 A, with a – sign that indicates that V2 is discharging.
And finally from equation 1,
I1 = (4.31 – 0.46) A
I1 = 3.85A
Hence the currents flowing through R1, R2, and R3 branches are 3.85, 4.31and 0.46A respectively.
Task 2 – Learning Outcome 3.2
Solve problems using circuit theorems to calculate currents and voltages in circuits
2) Three resistors of 7Ω, 12Ω, and 14Ω respectively are connected in parallel. This combination is connected in series with another parallel combination of 16Ω and 19Ω. If the whole circuit is supplied from a 24V source, calculate:
a. The total resistance.
b. The total current.
c. The PD across the 12Ω resistor.
d. The current through the 19Ω resistor.
Solution:
As per the given condition considering the fig. 3.2.1;
a: For total resistance, Rtotal, let if R6 is equivalent resistances for branch one comprising R1, R2 and R3 resistors then,
1/R6= 1/R1 +1/R2+ 1/R3

1/R6= 1/7 +1/12+ 1/14
1/R6= (12 +7+ 6)/84 = 25/84
R6= 3.36 Ω … (1)
Also, if R7 is equivalent resistance for R4 and R5,
1/R7= 1/R4 +1/R5
1/R7= 1/16 +1/19
1/R7= (19 +16)/304 = 35/304
R7= 8.685Ω …. (2)
Hence Rtotal = R6 + R7
Rtotal = 3.36 + 8.685
Rtotal = 12.045
b: For Itotal,
Itotal = V/Rtotal