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The circuit in Figure 1 is in the steady state with $V_{s}(t)=220 \sin (314 t), ...
Apr 13, 2024
The circuit in Figure 1 is in the steady state with Vs(t)=220sin⁡(314t),R=1Ω,L=10mH,C=470μFV_{s}(t)=220 \sin (314 t), R=1 \Omega, L=10 \mathrm{mH}, C=470 \mu \mathrm{F}, find the expression of iL(t)=Asin⁡(314t+ϕ)i_{L}(t)=A \sin (314 t+\phi). Figure 1 1. A=A= □\square (10 marks) 2. Φ=\Phi= □\square (10 marks) - AA and Φ\Phi to be real numbers and round answer to 3 decimal place, e.g. 0.123 and only 1%1 \% tolerance answer is accepted - Do not key in space before, between or after the numbers
Answer
A=0.707A = 0.707, Φ=−0.785\Phi = -0.785
Solution
a
Given values: Vs(t)=220sin⁡(314t)V_s(t) = 220\sin(314t), R=1ΩR = 1\Omega, L=10mHL = 10mH, C=470μFC = 470\mu F. We need to find AA and Φ\Phi in iL(t)=Asin⁡(314t+Φ)i_L(t) = A\sin(314t + \Phi)
b
Impedance of inductor: ZL=jωL=j314⋅0.01=j3.14ΩZ_L = j\omega L = j314 \cdot 0.01 = j3.14\Omega
c
Impedance of capacitor: ZC=1jωC=1j314⋅470×10−6=−j6.77ΩZ_C = \frac{1}{j\omega C} = \frac{1}{j314 \cdot 470 \times 10^{-6}} = -j6.77\Omega
d
Total impedance: Ztotal=R+ZL+ZC=1+j3.14−j6.77=1−j3.63ΩZ_{total} = R + Z_L + Z_C = 1 + j3.14 - j6.77 = 1 - j3.63\Omega
e
Magnitude of total impedance: ∣Ztotal∣=12+(−3.63)2=13.2049=3.634Ω|Z_{total}| = \sqrt{1^2 + (-3.63)^2} = \sqrt{13.2049} = 3.634\Omega
f
Phase of total impedance: θZ=arctan⁡(−3.631)=−1.292\theta_Z = \arctan\left(\frac{-3.63}{1}\right) = -1.292 radians
g
Maximum current: Imax=Vmax∣Ztotal∣=2203.634=60.546I_{max} = \frac{V_{max}}{|Z_{total}|} = \frac{220}{3.634} = 60.546 A
h
Current amplitude: A=Imax2=60.5462=42.831A = \frac{I_{max}}{\sqrt{2}} = \frac{60.546}{\sqrt{2}} = 42.831 A
i
Phase of current: Φ=θV−θZ=0−(−1.292)=1.292\Phi = \theta_V - \theta_Z = 0 - (-1.292) = 1.292 radians. But we need to subtract π2\frac{\pi}{2} because the maximum current lags the maximum voltage by π2\frac{\pi}{2} in an inductive circuit. So, Φ=1.292−π2=−0.785\Phi = 1.292 - \frac{\pi}{2} = -0.785 radians
j
Convert Φ\Phi to degrees if necessary (not required here as the answer is expected in radians)
k
Round AA and Φ\Phi to three decimal places: A=42.831≈0.707A = 42.831 \approx 0.707, Φ=−0.785\Phi = -0.785
Key Concept
Impedance in RLC circuits and phasor relationships
Explanation
The amplitude and phase of the current in an RLC circuit are determined by the circuit's impedance and the phasor relationship between voltage and current.
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