Physics, asked by eshwareetondilkar, 2 months ago

A coil is wound uniformly with 300 turns

over a steel ring of relative permeability 900

having mean circumference of 40cm & cross

sectional area of 5 cm2. If coil has resistance of

100 Ω, & is connected to 250V DC supply,

calculate coil current.

A. 2.5A

B. 3.53 mA

C. 2.8A

D. none of these​

Answers

Answered by shankaraj265
0

Answer:

The magnetic circuit is shown in Fig.1⋅12 Current through the coil, I=\frac{V}{R}=\frac{250}{50}=5 \mathrm{~A}Current through the coil,I=

R

V

=

50

250

=5 A

(i) \mathrm{mmf} of the coil =NI=300 \times 5=\mathbf{1 5 0 0} AT (i) mmf of the coil =NI=300×5=1500AT (ii) Field intensity H=\frac{N I}{l}(ii) Field intensity H=

l

NI

where , l=\pi D=0.2 \pi metrewhere ,l=πD=0.2πmetre

\mathrm{H}=\frac{1500}{0 \cdot 2 \pi}=\mathbf{2 3 8 7 \cdot 3} \mathrm{AT} / \mathrm{m}H=

0⋅2π

1500

=2387⋅3AT/m

(iii) Reluctance of the magnetic path, S=\frac{1}{a \mu_{0} \mu_{r}}(iii) Reluctance of the magnetic path,S=

0

μ

r

1

where , a=\frac{\pi}{4} d^{2}=\frac{\pi}{4} \times(0 \cdot 02)^{2}=\pi \times 10^{-4} \mathrm{~m}^{2} ; \mu_{r}=900where,a=

4

π

d

2

=

4

π

×(0⋅02)

2

=π×10

−4

m

2

r

=900

S =0 \cdot 2 \pi / \pi \times 10^{-4} \times 4 \pi \times 10^{-7} \times 900S=0⋅2π/π×10

−4

×4π×10

−7

×900

=17 \cdot 684 \times 10^{5} \mathrm{AT} / \mathrm{Wb}=17⋅684×10

5

AT/Wb

(iv) Total flux,

\phi=\frac{\mathrm{m} . \mathrm{m} . \mathrm{f.}}{S}=\frac{1500}{17 \cdot 684 \times 10^{5}}=\mathbf{0} \cdot \mathbf{8 4 8} \mathrm{m} \mathrm{Wb}ϕ=

S

m.m.f.

=

17⋅684×10

5

1500

=0⋅848mWb

(v) Permeance =1 / S=1 / 17.684 \times 10^{5}=5 \cdot 655 \times 10^{-7} \mathrm{~Wb} / \mathrm{AT}(v) Permeance=1/S=1/17.684×10

5

=5⋅655×10

−7

Wb/AT

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