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
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=
aμ
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
;μ
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