During the positive half cycle of the supply, diodes D1 and D2 conduct are forward biased and conduct current while diodes D3 and D4 are reverse biased and they act as an open circuit, the current flows through the load.
The resistance R3 & R4 are connected in series = (30 + 70) = 100Ω Now three resistor i.e 100Ω, 100Ω & 50Ω is connected in parallel 1/Rp = 1/100 + 1/100 + 1/50 Rp = 100/4 = 25 Ω Resistance 50Ω and 25Ω are connected in series. Rtotal = (50 + 25)Ω Rtotal = 75Ω
The inductance of the coil is given by the relation L = Nφ/I Where N = number of turns = 50 φ = flux = 200μWb I = current = 8 A L = 50 × 200 × 10−6 ⁄ 8 L = 1.25 mH
Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φA No. of turns NA = NB = 1000 Flux linkage in a coil with B = Flux linkage in coil A × 80/100 = 0.8 × 5 × 10−5 = 4 × 10−5 wb =0.04 mwb
Magnetic Field Strength (H) gives the quantitative measure of strongness or weakness of the magnetic field.
H = B/μo
Where
B = Magnetic Flux Density
μo = Vacuum Permeability
Magnetic Field strength at the center of circular loop carrying current I is given by
B = μoI/2r
B/μo = I/2r
H = I/2r
Where r = Radius
Now Given Parameters
Diameter = 1m
Current = 2A
∴ Magnetic field Intensity H = (2 / 2 × 1/2) = 2 A/m
EXAMIANS
