Equivalent inductance problems

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  Since we are given the voltage and current waveforms we can find the equivalent inductance of the circuit = 25 / 0.014 = 1,786 H.

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  R2 and R3 are effectively in parallel giving an equivalent resistance = (4 x 8)/12 = 2.67 ohm. This in turn is in series with R1 giving a total equivalent resistance across battery = 2.67 + 6 = 8.67 ohm.

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  Hence, the equivalent inductance in the circuit = L = 4H V = L dI/dt Or I = 1/L∫V dt = 6/4 ∫cos(100t) dt = 6/(100x4) sin(100t) = 0.015 sin(100t) Answer : I = 0.015 sin(100t) A Problem 3

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  Equivalent impedance seen by source = 100 + Zp = 114.1 -j56.4 = 127.28 @ -26.3 deg. Source current = Is = 20/127.28 = 0.16 A @ 26.3 deg. Hence Il, the current through the inductive load = = Is * Zc / (Zc+Zs) = 0.09 @66.9 deg.

• NEXT and FEXT crosstalk calculations

  We can evaluate these ratios since we are given the unit length mutual inductance between the lines as and the unit length line inductance as so In addition as we are given the unit length mutual capacitance between the lines as and the