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3 years ago

How to solve circuit theory using mesh analysis

Engineering

1 answer:

RoseWind [281]3 years ago

3 0

Explanation:

Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.

This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.

<u>Example</u>

Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...

mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)

mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs

mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0

You will note that the matrix of equation coefficients is symmetric.

In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.

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A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu

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Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

$emf= N\frac{d \phi}{dt}$

where;

ΔФ is change in magnetic flux

ΔФ = BA = B x πr²

ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²

$emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V$

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

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