Figure 6 shows the voltage phasors at a certain time of a...

Figure 6 shows the voltage phasors at a certain time of a circuit containing an ideal 100 ?F capacitor, an 80 mH inductor, a 20.0 ? resistor. The circuit is connected to an alternating current supply i(t)=Imaxsin(?t-?). Here the frequency of the alternating current supply is f = 60 Hz. VL = 60V VR = 50V VC = 30 V Figure 6 i) Draw the correct RLC circuit diagram used to obtain the voltage phasor diagram. ii) With reference to Figure 6, determine the sum of the voltage Vm of the RLC circuit. iii) Calculate the capacitive reactance, inductive reactance, and hence the impedance of the circuit. iv) Calculate the current Im and the value of phase angle ? of the circuit. v) Is the circuit is in resonance? Explain your answer.

Transcript

00:02 The resistance the inductance and the capacitance they share us same current it means they are connected in series right so this is the voltage source so this is our required circuit mario student this is resistance this is inductor and this is capacitor right and this is our voltage source all right in part two we have to the total voltage all right that would simply be equal to the voltage across the resistance squared plus the voltage across the inductor minus the voltage across the capacitor hole square this is the total voltage of the circuit that's would simply be equal to let's plug in the values voltage across resistor is given there is 50 volts square plus voltage across the inductor is given there is 60 volts minus voltage across the capacitor is given there is 30 volts square so from here the voltage in the total circuit due to all these components equal 10 multiplied by square root of 34 volts in the third part my student we have to find the capacitive reactance that would simply be equal to capacity reactance is one device by the angular frequency times the capacitance let's plug in the values one divided by angular frequency is given there is three hundred and seventy seven radiance per second multiplied by ten raised to the power two capacitor is more ten rest to the power minus six ferrad so from here the value of capacity reactants equals 6 .53 oms.

02:20 Likewise the value of the inductive reactants should be equal to should be equal to omega times l or that is a 2 pi times l that would simply be equal to omega is basically 2 pi and 2 pi f times l so let's plug in the values my dear student this equals 377 multiplied by the value of the l is 80 millie.

03:15 From here it equals the inductive reactants equal 30 .1 oom.

03:22 Right.

03:24 The impedance, the form of the appearance of the circuit is simple.

03:28 There is the square of the resistive load minus the difference of the inductive and the capacity load holds scared...

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