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A Parallel RLC Circuit
A resistance R, inductance L and capacitance C are connected in parallel across an AC source whose time- varying voltage is given by v(t) = V cos wt. Let R = 333 , L = 0:01 H, and C = 0:32 F. Also assume that V = 100 V and w= 12;500 rad/s. Note that in this circuit, the voltage is common to all three... 顯示更多 A resistance R, inductance L and capacitance C are connected in parallel across an AC source whose time- varying voltage is given by v(t) = V cos wt. Let R = 333 , L = 0:01 H, and C = 0:32 F. Also assume that V = 100 V and w= 12;500 rad/s. Note that in this circuit, the voltage is common to all three elements, and it is the currents through each element that add to give the total current delivered by the AC source. a) What are the peak currents Ir, IL, and Ic through the resistor, inductor, and capacitor? b) Draw a phasor diagram showing the current phasors Ir, IL, and IC, with the source voltage phasor V aligned with the positive real axis. (Make sure IR, IL, and IC are to scale with respect to each other.) c) What is the peak current delivered by the AC source? d) What is the phase angle between the source current and the source voltage? Does the source current lead or lag the source voltage?
最佳解答:
I think the value of capacitance given in the question should be 0.32 uF instead of 0.32 F. A capacitor of 0.32 F would give a tremendously large current, which is physically impossible. (a) Ir = 100/333 A = 0.3 A IL = 100/(12500 x 0.01) A = 0.8 A Ic = 100 x 12500 x 0.32x10^-6 A = 0.4 A (b) Ir aligns on the +x axis, same as V; Ic aligns on the +y axis; and IL aligns on the -y axis. (c) Peak current Io is, (Io)^2 = (IL - Ic)^2 + (Ir)^2 i.e. Io = square-root[(0.8 - 0.4)^2 + 0.3^2] = 0.5 A (d) Phase angle = arc-tan(0.4/0.3) = 53.13 degrees (= 0.927 radians) The source current lags behind the source voltage.
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A Parallel RLC Circuit
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發問:A resistance R, inductance L and capacitance C are connected in parallel across an AC source whose time- varying voltage is given by v(t) = V cos wt. Let R = 333 , L = 0:01 H, and C = 0:32 F. Also assume that V = 100 V and w= 12;500 rad/s. Note that in this circuit, the voltage is common to all three... 顯示更多 A resistance R, inductance L and capacitance C are connected in parallel across an AC source whose time- varying voltage is given by v(t) = V cos wt. Let R = 333 , L = 0:01 H, and C = 0:32 F. Also assume that V = 100 V and w= 12;500 rad/s. Note that in this circuit, the voltage is common to all three elements, and it is the currents through each element that add to give the total current delivered by the AC source. a) What are the peak currents Ir, IL, and Ic through the resistor, inductor, and capacitor? b) Draw a phasor diagram showing the current phasors Ir, IL, and IC, with the source voltage phasor V aligned with the positive real axis. (Make sure IR, IL, and IC are to scale with respect to each other.) c) What is the peak current delivered by the AC source? d) What is the phase angle between the source current and the source voltage? Does the source current lead or lag the source voltage?
最佳解答:
I think the value of capacitance given in the question should be 0.32 uF instead of 0.32 F. A capacitor of 0.32 F would give a tremendously large current, which is physically impossible. (a) Ir = 100/333 A = 0.3 A IL = 100/(12500 x 0.01) A = 0.8 A Ic = 100 x 12500 x 0.32x10^-6 A = 0.4 A (b) Ir aligns on the +x axis, same as V; Ic aligns on the +y axis; and IL aligns on the -y axis. (c) Peak current Io is, (Io)^2 = (IL - Ic)^2 + (Ir)^2 i.e. Io = square-root[(0.8 - 0.4)^2 + 0.3^2] = 0.5 A (d) Phase angle = arc-tan(0.4/0.3) = 53.13 degrees (= 0.927 radians) The source current lags behind the source voltage.
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