R = 35 Ω R=35\Omega R = 35Ω ,L = 180 m H , V = 100 V , f = 50 H z L=180mH,V=100V,f=50Hz L = 180 m H , V = 100 V , f = 50 Hz
a)Impedance (Z) of the coil
R = 35 Ω R=35\Omega R = 35Ω
X L = 2 π f L = 2 π × 50 × 0.18 = 56.6 Ω X_L=2\pi fL=2\pi ×50×0.18=56.6\Omega X L = 2 π f L = 2 π × 50 × 0.18 = 56.6Ω
Z = R 2 + X L 2 = 3 5 2 + 56. 6 2 = 66.5 Ω Z=\sqrt{R^2+X_L^2}=\sqrt{35^2+56.6^2}=66.5\Omega Z = R 2 + X L 2 = 3 5 2 + 56. 6 2 = 66.5Ω
b)Current (I) consumed by the coil
V = I × Z V=I×Z V = I × Z
∴ I = V Z = 100 66.5 = 1.5 A \therefore I=\frac{V}{Z}=\frac{100}{66.5}=1.5A ∴ I = Z V = 66.5 100 = 1.5 A
c)The power factor and phase angle,Φ \Phi Φ
C o s ϕ Cos\phi C os ϕ = R Z o r S i n ϕ = X L Z o r t a n ϕ = X L R =\frac{R}{Z}or Sin\phi =\frac{X_L}{Z} or tan\phi =\frac{X_L}{R} = Z R or S in ϕ = Z X L or t an ϕ = R X L
∴ C o s ϕ = R Z = 35 66.5 = 0.5263 \therefore Cos\phi =\frac{R}{Z}=\frac{35}{66.5}=0.5263 ∴ C os ϕ = Z R = 66.5 35 = 0.5263
C o s − 1 ( 0.5263 ) = 58.2 ° Cos^{-1}(0.5263)=58.2° C o s − 1 ( 0.5263 ) = 58.2°
d)Apparent power (S) consumed by the coil
P = V × I c o s ϕ = 100 × 1.5 c o s ( 58.2 ° ) = 79 W P=V×I cos\phi =100×1.5cos(58.2°)=79W P = V × I cos ϕ = 100 × 1.5 cos ( 58.2° ) = 79 W
Q = V × I s i n ϕ = 100 × 1.5 × s i n ( 58.2 ° ) = 127.5 V A r Q=V×Isin\phi =100×1.5×sin(58.2°)=127.5VAr Q = V × I s in ϕ = 100 × 1.5 × s in ( 58.2° ) = 127.5 V A r
S = V × I = 100 × 1.5 = 150 V A S=V×I=100×1.5=150VA S = V × I = 100 × 1.5 = 150 V A
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