Including mutual inductance for the two parallel wire structure above the follow

Question

Including mutual inductance for the two parallel wire structure above the following figure, determine how the magnitude of the interference voltage, V in wire R depends ground, shown in the on hc h results for two frequencies: 2 MHz and 5 MHz of the source connected to wire G. The source has its magnitude of 10 volts. The length of each wire is 1meter. us when the vertical position of wire G, ho changes from 1 em to 6 cm. Calculate the Plot the relationship between Vs and ho for both frequencies. 3cm UR ng = 3cm h P Two wiRS A BovE THE GROUND Rs = RL-50 SZ Vi

Explanation / Answer

Vn in wire R depends on Hg because of mutual inductance between two wires. As Hg changes from 1cm to 6cm its length from ground increases so the voltage generated in another wire also changes due to mutual inductance.

As shown, Rs = Rl = 50ohm, the source voltage is already mentioned in the above i.e., 10V. So, calculate current through the wire G, i.e.,

Vs = I(Rs + Rl)

10 = I(50 +50)

10 = I(100)

I = 10/100

I = 1/10 = 0.1amp

So, let us assume that same current flows through the wire R due to mutual inductance,

Vn = I(100K +100K) * Hg where, Hg = Hg1 - Hg2 = 6-1 = 5cm

= 0.1(200K) *5

= 20K * 5

.: Vn =100KV

For 2MHz,

Vg = Vs * sin(2 * 3.14 * f *t)

Vg = 10 * sin(6.28 *2M * 1)

Vg = 10* sin(12.56 * 1000000)

Vg = 10 * (-0.64278)

Vg = 6.4278V

For 5MHz,

Vg = 10 *sin(2*3.14*5M*1)

Vg= 10*sin(31400000)

Vg = 10* 0.98480

Vg = 9.848V

So, the Vn generated is related to hg as hg changes vn also changes.

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