A flux of 0.0006Wb is required in the air – gap of an iron ring of cross – section 5.0 cm2and mean length 2.7 m with an air – gap of 4.5 mm. Determine the ampere turns required. Six H values and corresponding B values are noted from the magnetization curve of iron and given below.
H(At/m) |
200 |
400 |
500 |
600 |
800 |
1000 |
B(Wb/m2) |
0.4 |
0.8 |
1.0 |
1.09 |
1.17 |
1.19 |
Given that, Flux=0.0006 Wb
Cross section area \(A=5\ {cm}^2=5\times{10}^{-4}m^2\)
The flux density \(B=\frac{0.0006}{5\times{10}^{-4}}=1.2\ Wb/m^2\)
From the given data, we have
\(H(AT/m)=\frac{1000AT}{m}forB=1.2Wb/m^2\)
Hence ampere turns required for the iron part producing the given flux density.
Flux density =\(Hl=1000\times2.7\ At=2700\ AT\)
Extra ampere turns necessary to produce the same flux in the air gap is
\(B=\mu H\)
Here \(H=\frac{B}{\mu_0}=\frac{1.2}{4\pi\times{10}^{-7}}=954930\ AT\)
Extra ampere turns required is \(954930\times4.5\times{10}^{-3}AT=4297.185\ AT\)
The total number of ampere turns necessary is \(\left(2700+4297.185\right)=6997.185\ AT\)
Neglecting leakage and fringing field.
Find VAB from the circuit if all the resistances are of same value of 1 Ω.
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Fig. 20(a)
For the circuit shown below, find the potential difference between a and d:
