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8/7/2019 elektrik di rumah

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I = 1.5A R1 = 8 R2 = 6

Vj R3 = 4

Figure A2.1

1. A2.1 based circuit diagram above, calculate;i. total resistance

Rj

= R1+R

2+R

3= 8+6+4 = 18

ii. value VjVj = IRj = (1.5)(18) = 27V

iii. voltage drop at R3, VR3( R3/Rj) Vj = (4/18) 27 = 6V

2. three resistors connected in parallel to each value R1 = 6, R2 = 5, R3 = 20 and issupplied with 100 V supply. Calculate

1. total resistance1/Rj = 1/R1 + 1/R2 + 1/R3 = 1/6 + 1/5 + 1/20 = 0.417

Rj = 1/0.417 = 2.4

2. total currentIj = V/Rj = 100/2.4 = 41.7 A


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3. voltage across each resistorV1 = V2 = V3 = V = 100V

4. current per resistorI1 = V/R = 100/6 = 16.7 AI2 = V/R = 100/5 = 20 AI3 = V/R = 100/20 = 5 A

3. please define for the first Kirchoff law and the second Kirchoff lawThe first Kirchoff law states that the amount of current to at one point is equal to the total

current leaving that point, or at any crossing point in the circuit, the total current enteringthe point algebra is equal to the total flow out

Second Kirchoff law states that in a closed circuit, the sum of the voltage rise and voltage

drop is zero, or in any electrical circuit is completed, the amount of algebra to increase the

voltage must equal the total voltage drop


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Based on 2.1 K circuit diagram below, find;

1. total resistanceR23 = R2 R3/R+R3

= (4)(8)/4+8= 2.667

Rj = R1 R23/R1+R23= (2)(2.667)/4.667= 1.143

2. voltage at R2V2 = Vs = 240 V

3. current at R2 and R3I2 = (Vs/R2) = (240/4) = 60AI3 = (Vs/R3) = (240/8) = 30A

4. power dissipated in R1 and the resistance of the whole circuitP1 = (Vs)

2/R1 = (240)2/2 = 28.8kWPj = (Vs)

2/Rj = (240)2/1.143 = 50.4 kW


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according to figure 2.3, calculate the current value of each branch and voltage drop in eachresistor using Kirchoff law.

At nod A : I1 = I2+I3-----------------(1)Gelung 1: -12 + 4I2 + 4 + (1)I1 = 0 ------------------ (2)

Gelung 2: -4-3I2 + 5I3 + 6 = 0

-3I2+5I3 = -2------------------------(3)

Substitute(1) into (2)

4I2 + (I3 + I2) = 8

5I2 + I3 = 8-------------------------(4)

solve the equation (3) and (4) using Cramer rule or other appropriate methods

- 3 5 I2 = -2 => D = -3 5

5 1 I3 8 5 1

determining the value of , D = -3 5 = (-3)(1)-(5)(5) = -28

5 1

I2 = -2 5 = (-2)(1)-(5)(8) = -42 I3 = -3 -2 = (-3)(8)-(-2)(5) = -14

8 1 5 8

I3 = I3/D = -14/-28 = 0.5A


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RL circuit is a 10 ohm series resistance and inductance 0.2H supplied with the supply of 250 VAC, 50Hz. calculate;

1. circuit impedance2. circuit current3. phase angle

solution:

Give R = 10, L = 0.2H, V = 250 V and f = 50Hz

Where, XL = 2fL = 2(50)(0.2) = 62.83

1) impedance, Z = R2 + XL2 = 102 + 62.832 = 63.622) circuit current, I = V/Z = 250/63.62 = 3.93A3) phase angle, tan-1 (XL/R) = tan-1(63.62/10) = tan-1(6.362) = 81.1

RL circuit is a 10 ohm series resistance and capacitive 200f supplied with the supply of 75VAC, 50Hz. calculate;

1. circuit impedance2. circuit current3. power factor

Give R = 10, C =200f, V = 75 V and f = 50Hz

Where, Xc = 1/2fC = 1/2(50)(200 x 10-6) = 15.92

1) impedance, Z = R2 + Xc2 = 102 + 15.922 = 18.82) circuit current, I = V/Z = 75/15.92 = 4.71A3) power factor, cos (R/Z) = cos(10/15.92) = 0.628


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9/10

1. give the definition of the impedanceimpedance is the number of barriers that exist within ac circuits

2. state one (1) the definition of power factorPower factor is the ratio of resistance to impedance

3. draw and label a series RL circuit in an ac circuit

4. a 200 micro Farad capacitor Capacitive connected to supply 75V, 50Hz. Capacitivereactance value and what is current in the circuit?

Xc = 1/2fC = 1/2(50)(200 x 10-6) = 15.92

circuit current, I = V/Z = 75/15.92 = 4.71A

5. draw and label a vector diagram for a series RLC circuit for Xc > XL


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