Question Part Points Submissions Used Using Kirchhoffs rules

Question Part

Points

Submissions Used

Using Kirchhoff\'s rules, find the following. (1 = 70.6 \', 2 = 61.2 V, and 3 = 78.6 v.) 4.00 k2 2 R 2.00 k (a) the current in each resistor shown in the figure above MA MA MA (b) the potential difference between points c and f magnitude of potential difference point at higher potential -Select Need Help?RoadiMastritTalk to a Tutor Read It Master It

Solution

For a closed circuit we know the following Kirchhoff\'s rules:

1.) Junction law: The amount of charge (hence current) flowing at a junction is same as the amount of charge (hence current) flowing out of it.

2.) Loop law: The net voltage drop across a closed circuit is zero.

We will use the above two laws to determine the values needed.

a.) Let us assume that the current in R1 flows from f to a, for R3 it flows from d to c and for R2, it flows from c to f.

Now using loop law for the circuit on the left, we have:

E1 - E2 = 3I_R2 + 2I_R1

or, 70.6 - 61.2 = 9.4 = 3I_R2 + 2I_R1

Also, for circuit on the right, we have:

E3 - E2 = 4I_R3 + 3I_R2

or, 17.4 = 4I_R3 + 3I_R2

Using junction law at f, we get: Ir2 = Ir1 + Ir3

Using in the equations above, we get:

9.4 = 3I_R3 + 5I_R1 (Eq 1)

and, 17.4 = 7I_R3 + 3I_R1 (Eq 2)

Multiply equation Eq 1 by 3 and Eq2 by 5, and then resolving, we get:

28.2 = 9Ir3 + 15 Ir1

87 = 35 Ir3 + 15 Ir1

or, 58.8 = 26 Ir3

or, Ir3 = 2.2615 mili-amperes

Putting in Eq 1, we get:

Ir1 = [9.4 - 3(2.2615 )] / 5 = 0.5231 mili-amperes

Ir2 = 0.5231 + 2.2615 = 2.7846 mili-amperes

b.) For point c and f, the current would be flowing from c to f.

Therefore the voltage drop while moving from c to f would be:

E2 - 3Ir2 = 61.2 - 3(2.7846) = 52.8462

Therefore the magnitude of the difference is 52.8462

Also, as the current flows from c to f, point c will be at higher potential.