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Q. Consider a pair of coupled coils as shown in Figure of the text, with currents, voltages, and polarity dots as indicated. Show that the mutual inductance is L12 = L21 = M by following these steps:
(a) Starting at time t0 with i1(t0) = i2(t0) = 0, maintain i2 = 0 and increase i1 until, at time t1, i1(t1) = I1 and i2(t1) = 0. Determine the energy accumulated during this time. Now maintaining i1 = I1, increase i2 until at time t2, i2(t2) = I2. Find the corresponding energy accumulated and the total energy stored at time t2.
(b) Repeat the process in the reverse order, allowing the currents to reach their ?nal values. Compare the expressions obtained for the total energy stored and obtain the desired result.
If a current of 20A flows for five minutes, find the quantity of electricity transferred. Quantity of electricity, Q = It = 20 x 5 x 60 = 6000C
Generation ofwide-band FM can be done by variousmeans.However, only themost common and conceptually the simplest one, known as the direct method, is considered here. It employs a
Q. The inductance per unit length in H/m for parallel-plate infinitely long conductors in air is given by L = µ 0 d/w = 4π×10 -7 d/w, where d and w are inmeters. Compute L (per un
Perform a Hartree-Fock geometry optimization calculation of butadiene using a minimal basis set. Repeat with the 6-311G(d,p) basis set, using the optimized minimal basis set geomet
If a current is passed through a coil, it creates a flux within the coil. Any attempt to change this flux will create a back emf that acts to oppose the change. Consider again
matlap code to solve the fast decoupled method
In this case, 24 simultaneous calls can be put through switch. Generally a 24-outlet Uni-selector is used as a selector hunter. Every one of the 24 outlets is connected to one two-
A single-phase, 50-kVA, 2400:240-V, 60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding R1 = 0.75 Resistance of the 240-V winding R2
Three transformers are provided with the following rating values: 1) 2400/240 V, 55kVA, Req1 = (0.1 + 0.0Y) Ohm and Xeq1 = (0.15 + 0.XX) Ohm. 2) 240/120 V, 25kVA, Req1 = (0.1
Explain junction transistors (npn and pnp). Junction Transistor: This transistor consists of two p-n junctions combined in one crystal as demonstrated in figure below.
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