**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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