Here\'s the question: The emf\'s and the resistors in the two-loop circuit are g

Question

Here's the question:

The emf's and the resistors in the two-loop circuit are given as indicated in the figure. The arrow directions in the circuit set up the sign convention. The current values can be positive or negative. Find I3 in terms of I1 and I2 by Kirchhoff's Rule I. Write down explicitly Kirchhoff's Rule II for the left loop, ONLY in terms of I1 and I2 (no symbol I3) Write down explicitly Kirchhoff's Rule II for the right loop, ONLY in terms of I1 and I2. (no symbol I3) Determine values of currents I1, I2, and I3. Find the voltage difference Va - Vb across points a and b.

Explanation / Answer

(a) Kirchoffs 1st (current law) states that the current leaving any node must be equal to the current entering that node

so I3 = (I1 + I2)

or I1 + I2 - I3 = 0

(b)Kirchoffs Volatge law states that the potential difference across any loop is zero

So starting from the 10V source and going clock wise, we have

-10 + 4 * I1 - 3 * I3 + 8 = 0

=> 4 * I1 - 3 * I3 = 2

now since the question asks for the equation to be in I1 and I2 only, we will employ the equation

I3 = I1 + I2

so 4 * I1 - 3 * (I1 + I2) = 2

=> I1 - 3 * I2 = 2

which is your required equation for the left loop

(c) Now for the right loop

-8 + 3 * I3 - I2 + 17 = 0

=> -8 + 3 * (I1 + I2) - I2 + 17 = 0

=> 3 * I1 + 3 * I2 - I2 + 9 = 0

=> 3 * I1 + 2 * I2 = -9

(d) Now solve the last two equation obtained

I1 = 2 + 3 * I2

replacing this in 3 * I1 + 2 * I2 = -9

3 * (2 + 3 * I2) + 2 * I2 = -9

=> 6 + 9 * I2 + 2 * I2 = -9

=> 11 * I2 = -9 -6 = -15

=> I2 = -15/11 = -1.3636 A

then I1 = 2 + 3 * -1.3636 = -2.0909 A

(e)

I3 = I1 + I2 = -3.4545 A

then p.d between a and b = 8 - 3 * I3 = 8 + 3 * 3.4545 = 18.3635 V

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---