Solutions _ Tutorial 12_revision.pdf

Solutions _ Tutorial 12_revision.pdf - Manora Caldera...

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Unformatted text preview: Manora Caldera Electrical Systems Tutorial 12 – Solutions 1. In this circuit, Applying KCL to the non-inverting node gives, 0 3 gives Applying voltage divider rule gives ∙ 1 gives 2. (a) ∙ Boolean Expression: Truth Table: ∙ 0 0 1 1 0 1 0 1 1 1 0 0 1 0 1 0 1 1 1 0 0 1 1 1 ∙ ∙ 0 1 1 1 0 1 1 0 (b) ∙ Boolean Expression: ∙ Truth Table: 0 0 1 1 0 1 0 1 1 1 0 0 1 0 1 0 1 0 0 0 ∙ 0 1 1 1 ∙ 0 1 1 1 1 Manora Caldera 3. Assuming the diode is ON, applying KCL to the nodes at the two ends of the diode gives, A B Applying KCL to Node (A), 0 10 30 Applying KCL to Node (B), 0 8 0 12 0 0 Also, when the diode is conducting, 0.7 and given that equations give, 1.4024 , 0.7024 146.3 As 0, the original assumption of the diode is ON is correct. 10 . Solving the 4. Assume the diode is ON and applying Mesh equations to the two independent loops, Mesh 1 (one on the left): 40 6 60 0 Mesh 2 (one on the right): 40 0.7 30 As give, 0 , solving the equations 31.48 As the current through the diode is positive, the assumption that the diode is ON is correct. Hence the current through the diode is 31.48mA. 2 Manora Caldera 150, 5. Given that 4 , 0.7 For the input loop (B-E loop), applying KVL gives 10 0 and As 4 , As 1 0.5 8 . 52.98 , 7.947 and Applying KVL to the output loop (C-E loop), 18 1 18 Therefore, 1 7.947 10 and, 10 4 6.053 52.98 10 0.7 4 5.23 6. For the input loop (B-E loop), applying KVL gives 5 10 5 10 0.4 0.7 0.4 1 . and hence 0.4 121 0 73.63 . 0.4 0 0.4 1 73.63 10 3.56 For the output loop (C-E loop), applying KVL gives 12 This gives 0.5 12 0.5 0.5 120 3.56 73.63 10 3.56 4.02 3 Manora Caldera 7. 50√2 As 50 50 rad/s , the angular frequency 50 . . 0.2 10Ω 10Ω 10 90°Ω . 10Ω 10Ω 10 90°Ω 2Ω 2Ω 2 90°Ω // As total impedance Hence, 10 3 2 // 8 10 3.22 . 10 1.07 10 3.22 11.07 Ω 11.53 73.78°Ω The current through the 2mF capacitor is the total current coming from the ac source. ° As . . 4.34 ° ° √ . . 6.13 In time domain, 73.78° in rms OR 6.13 ° 73.78° in peak 50 73.78° Voltage drop across 2mF, 6.13 Current through 73.78° 10 90° 61.3 16.22° can be found out using current divider rule, . . 6.13 0.94 73.78° 15.36° 6.13 73.78° 0.91 0.25 6.13 5.76 89.14° 73.78° Voltage drop across 3, 5.76 89.14° 3 17.28 89.14° Voltage drop across 10mF, 5.76 Similarly, current through . . 89.14° 2 90° 11.52 0.86° can be found out using current divider rule, . 4 Manora Caldera 6.13 . 0.266 . 73.78° 70.2° 6.13 0.09 73.78° 0.25 6.13 1.66 73.78° 3.58° Voltage drop across 8, . 1.66 3.58° 8 13.28 3.58° 1.66 3.58° 10 90° 16.6 Voltage drop across 0.2H, . . . 93.58° From Figure 8(b), Period T of V2 = 1 sec. 8. Hence frequency the angular frequency 2 2 1 and π 1 6.283 rad/s Peak amplitude of V2 = 2 volts, and the phase shift is zero. Thus, 2 sin 6.283 √ 0° 1.414 1 Time shift between V2 and V1= Δ 0° in rms 0.2 and 5 . 360° The phase shift between V2 and V1 360° 72° Thus, the Peak amplitude of V1 = 8 volt, and the phase shift is 72° lagging. Hence, 8 sin 6.283 72° 72° 5.66 72° in rms √ (i) In phasor domain = (ii) . Current through the resistor R = . √ 0° 0.141 5.66 0.335 Voltage across Impedance . . . ° 0.2 sin 6.283 and 0° in rms 72° 1.414 0° 5.383 5.393 86.44° ° 38.25 86.44° 2.38 38.18 Ω Thus, Z consists of a resistor of 2.38Ω and a capacitive reactance of XC= 38.18 Ω. 5 ...
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