Resposta:
The circuit diagram is shown in Fig. 15.16(a). Since the total circuit resistance is 8+5+108+5+10, i.e. 23\, \Omega23Ω, an equivalent circuit diagram may be drawn as shown in Fig. 15.16(b).
Inductive reactance,
X_{L} = 2\pi fLXL=2πfL
= 2\pi (20 \times 10^{3})(130 \times 10^{-6})=16.34\, \Omega=2π(20×103)(130×10−6)=16.34Ω
Capacitive reactance,
X_{C} = \frac{1}{2\pi fC}=\frac{1}{2\pi (20 \times 10^{3})(0.25 \times 10^{-6})}XC=2πfC1=2π(20×103)(0.25×10−6)1
= 31.83\, \Omega=31.83Ω
Since X_{C} > X_{L}XC>XL, the circuit is capacitive (see phasor diagram in Fig. 15.12(c)).
X_{C} \, -\, X_{L} = 31.83\, -\, 16.34 = 15.49\, \OmegaXC−XL=31.83−16.34=15.49Ω
(a) Circuit impedance, Z = \sqrt{R^{2} + (X_{C}\, -\, X_{L})^{2}} = \sqrt{23^{2} + 15.49^{2}} = 27.73\, \OmegaZ=R2+(XC−XL)2=232+15.492=27.73Ω
Circuit current, I= V/Z= 40/27.73=1.442\, AI=V/Z=40/27.73=1.442A
From Fig. 15.12(c), circuit phase angle
\phi = \tan^{-1}\, (\frac{X_{C}\, -\, X_{L}}{R})ϕ=tan−1(RXC−XL)
i.e.
\phi = \arctan^{-1}\, (\frac{15.49}{23})=33.96^{\circ}\: \textbf{leading}ϕ=arctan−1(2315.49)=33.96∘leading
(b) From Fig. 15.16(a),
V_{1} = IR_{1} = (1.442)(8) = 11.54\, VV1=IR1=(1.442)(8)=11.54V
V_{2} = IZ_{2} = I\sqrt{5^{2}+16.34^{2}}V2=IZ2=I52+16.342
=(1.442)(17.09) = 24.64\, V=(1.442)(17.09)=24.64V V_{3} = IZ_{3}=I\sqrt{10^{2}+31.83^{2}}V3=IZ3=I102+31.832 =(1.442)(33.36) = 48.11\, V=(1.442)(33.36)=48.11V
The 40\, V40V supply voltage is the phasor sum of V_{1},\, V_{2}V1,V2 and V_{3}V3
