Use Mesh Analysis to get the current of each load. I, j60 2 20 Ω 80 Ω ww 100/120° V -j40 Q -j40 Q 60 -30° V
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A: Given
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A: We will use parallel and series combination of resistance to find out equivalent impedance.
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Q: 10 10 Find the equivalent impedance Zeq as seen from the given terminals. A 10+10j B 10-10j 10…
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Q: 2. Solve I using Nodal analysis 5/0° A 2Ω j1Q 20/-90° V (+ -j2 Q 21 4Ω
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Q: Using mesh analysis solve for 1 12 13 14 5+js 3-j3 3-j3 201
A: Mesh analysis uses Kirchhoff's voltage law and ohm's law. The detailed solution is provided below.
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A: The solution can be achieved as follows.
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A: The solution can be achieved as follows.
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Q: Find the total impedance of the circuit below. w- R = 22 Ž R = 10 2 100 V C = 180µF 50Hz. L= 33mH O…
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- 0 1540 in! 2+14 3-j8 500 71₂24+ [7] 24+514-52 Find the complex power absorbed by the load.A load consists of XL=30 Q and Xc=120 Q are connected in parallel across 120 Volt supply. The reactive power of the circuit is Reactive Power VARCalculate the currents 1₁ and 12 in phasor form using the current divider rule. I = 20 AZ40°¹ RM22N XL 60
- Get the equivalent resistance in point a and b b 62 +j85 Q 62+j85 Q a 12 - 109 Q 52-j3 Q 52-13 Q 12-j89 Q 12 - 1690Vdc across the load is 4:1 110 V rms R2 1.0k 12.36V O 24 73V O 110V O 55VA load consists of XL=55 and XC=100 are connected in parallel across 155 Volt supply. The reactive power of the circuit is Reactive Power VAR
- Using the current sivider rule fin the current through each impedance R x2 = 44X²622² I 20/01 3x2 31 R52Here R = 5 ohm C = -j 20 ohm L = j 20 ohm So find Zequivelent if all these element are in parallel. I will thumbsupFor the mutual electric circuit shown in figure beside, find the set values of mesh currents 1 & 12? * K=1 J4N J92 32 K=0.75 5Ω 10020 Volt I1 J42 12 -J92 -J92 O i1=22.1+j7.4 A 12=5.96+j0.01 A O i1=22.1-j7.4 A i2=5.96-j0.01A O i1=22.1+j7.4 A i2=5.96-j0.01 A O i1=22.1-j7.4 A i2=5.96+j0.01 A
- Proslem #6 Gluen ths ForLOwing Sot of $QUATIONS, CONSTMUET the Fouoaing GnaviT. V, -Vz 10 - j20 10 Vz Ve -Ve jie j to 01 Vg -V. 10 -j20 Vg - Ve j to - 20Question 14 Using the Node Voltage Method, what is the voltage V,? --j2 0 ww 4430° A Va VD j5 0 A 12430° V 12-30° A 20.71264.3° V O 10.82443.9° V O 11.22/83.9° V O 12.74Z - 83.7° VVery urgent. Draw the waveforms of the voltage at the load (RL) end, the voltage at the R end, the voltage at the L end and the current 1 when the switch K is turned to position 2 after a long enough time.