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SEJ102_Week 4 (Studio Solutions) 6 EXAMPLE 10-2 The current in figure is 0.2mA. Determine the source voltage and the phase angle. Draw the impedance triangle = = 0.2 × 18.8Ω = 3.76 The phase angle, = tan−1 = �2 + 2 = �(10 )2 + (15.9 )2 = 18.8 Ω = tan−1 � 15.9Ω 10 Ω � = 57.8° R 57.8 Z Xc 1 2 1 = Ω 2 × 1 × 0.01 = 15.9Ω = 7 EXAMPLE 10- 3 Determine the current in the RC circuit of Figure 10 = 5.30 = 1.89 = = �2 + 2 = �(2.2 )2 + (4.82 )2 = 5.30 Ω = 1 2 1 Ω 2 × 1.5 × 0.022 = 4.82Ω = 8 EXAMPLE 10-4 Determine the source voltage and the phase angle in Figure. Draw the voltage phasor diagram. Since VR and VC are 90° out of phase, you cannot add them directly. The source voltage is the phasor sum of VR and VC. = �2 + 2 = �(10 )2 + (15 )2 = 18 The phase angle between the resistor voltage and source voltage is 15 = tan−1 = tan−1 = 56.3° VR 10 56.3 VC VS 9 EXAMPLE 10-6 Determine the amount of phase lag from the input to output in the lag circuit in Figure. The phase lab between the output voltage and input voltage is Capacitive reactance = 1 2 = = 90° − tan−1 1 Ω 2(1 )(0.1) = 1.59 Ω, so = 90° − tan−1 = 23.2° The output voltage lags the input by 23.20 1.59 680 10 EXAMPLE 10-10 Determine the total admittance in Figure, and then convert it to impedance. To determine admittance, Y, first calculate the values for G and BC. The capacitive reactance is = The capacitive susceptance is = 1 1 = = 3.03 330 1 1 = = 723Ω 2 2 × 1 × 0.22 = 1 1 = = 1.38 723 Total admittance, = � 2 + 2 = �(3.03 )2 + (1.38 )2 = 3.33 1 1 Impedance = = 3.33 = 300Ω 11 EXAMPLE 10-13 Convert the parallel circuit in Figure to an equivalent series form. The equivalent series values are = cos () = sin Now, = 1 = � 2 + 2 1 1 = = 5.56 180 1 1 , = = = 3.70 270 = , = �(5.56 )2 + (3.70 )2 = 6.68 Equivalent series values = 1 1 = = 150Ω 6.68 = 150 × cos 33.6° = 125Ω () = 150 × sin 33.6° = 83Ω Example 11-1: Determine the inductance of the coil in Figure. The permeability of the core is 0.25 × 10−3 /. Solution: Inductance, = 2 where N is number of turns, A is area of the core and l is length. We have, l=1.5cm=0.015m, d=0.5cm=0.005m 2 0.005 2 = = � � = � � = 1.96 × 10−5 2 2 2 2 So, = (350)2 �0.25× 10−3 ��1.96×10−5 2 � 0.015 = 40 Example 11-2: Determine the total inductance for each of the series connections Solution a. = 1 + 2 + 1.5 + 5 = 9.5 b. = 5 + 2 + 10 + 1 = 18 Example 11-3: Determine LT Solution = 1 1 1 = = 1 1 1 1 1 1 0.8 1 + 2 + 3 10 + 5 + 2 = 1.25 Example 11-5: Calculate the RL time constant for the bellow circuit. Then determine current and the time at each timeconstant interval, measure from the instant the switch is closed. Solution The RL time constant is = The final current is 10 = = 8.33 1.2Ω = 12 = = 10 1.2Ω Using the time-constant percent value At 1: = 0.63(10) = 6.3; = 8.33 At 2: = 0.86(10) = 8.6; = 16.7 At 3: = 0.95(10) = 9.5; = 25.0 At 4: = 0.98(10) = 9.8; = 33.3 At 5: = 0.99(10) = 9.9 ≅ 10; = 41.7 Example 11-7: a. the circuit below has a square wave input. What is the highest frequency that can be used and still observe the complete waveform across the inductor? b. assume the generator is set to the frequency determined in (a). Describe the voltage waveform across the resistor? Solution: a. the period needs to be 10 times the time-constant to observe the entire wave. So ℎ , = 1 10 = = 15 = = 0.454 33 Ω 1 = 220 10 × 0.454 b. voltage across resistor will have same shape as the current wave form. Example 11-8: determine the inductor current 30 after the switch is closed Solution: = (1 − − ) Where is final current and is time constant The final current = = The RL time constant = 12 2.2 Ω = = 5.45 and 100 2.2 So 30 = (5.45 × �1 − − = 45.5 30 45.5 � = 2.63 Example 11-10: a sinusoidal voltage is applied to the circuit. The frequency is 10 kHz. Determine the inductive reactance. Solution: Inductive reactance, = 2 = 2 × 10 × 5 = 2(10 × 103 × 5 × 10−3 ) = 314 Ω Example 11-12: determine the rms current in below circuit Solution: = = 2 = 2(10 × 103 × 100 × 10−3 ) = 6283 Ω So, = 5 6283 Ω = 796 Example 12-2: the current in the below circuit is 200. Determine the source voltage. Solution: Applying ohm’s law, = The impedance = � 2 + 2 and Inductive reactance = 2 = 2(10 × 103 × 100 × 10−3 ) = 6.28Ω So, = �(10 Ω)2 + (6.28 Ω)2 = 11.8 Ω Now, = 200 × 11.8 Ω = 2.36 Example 12-3: determine the source voltage and the phase angel in below circuit. Draw the voltage phasor diagram. Solution: Since are 90° out of phase, we cannot add them directly. The source voltage is the phasor sum of . = �2 + 2 = �502 + 352 = 61 The phase angle between the resistor voltage and source voltage is 35 = tan−1( ) = 35° 50