Network Theorems (Part I)
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Network Theorems (Part I)-Numerical Problems Key points: - The problems considered in this set are involving both dependent and independent sources. Following points may be noted ο· Dependent sources are voltage or current sources whose output is function of another parameter in the circuit. ο· Dependent sources only produce a voltage or current when an independent voltage or current source is in the circuit. ο· Dependent sources are treated like independent sources when using nodal or mesh analysis but not with superposition. 1. In the following circuit find the value of vTH and RTH isc 2 2 8V 6 DC vTH, RTH π π½ππ = (π+π) × π = ππ½ Ans. (Replacing voltage source by it’s internal impedance) πΉππ = (πβπ) + π = π. πβ¦ 2. In the following circuit find the value of iN and RN 2 15V DC v1 2 4 iN, RN πΉπ΅ = πβπ + π Ans. = ππ π β¦ ππ΅ = short circuit current (πππ ), By source transformation the equivalent circuit is 2 isc 2 7.5A πππ = π.π×(πβπ) π+πβπ = π.π×π ππ π( ) π 4 ππ = ππ = π ππππ 3. In the following circuit find the value of vTH and RTH 2 2A Ans. vTH, RTH 1 3 π×π Current through 1β¦ resistor (using current division) = π+π = ππππ π½ππ = π ππππ πΉππ = πβπ = π×π π π = πβ¦ (Replacing current source by it’s internal impedance i.e. open circuit) 4. For the following circuit find the value of vTH and RTH 30 25 P 20 5V vTH, RTH DC 5A Ans. To calculate πΉππ , deactivate all independent sources i.e, replace them by their internal impedances. 30 25 RTH πΉππ = ππ + ππ = ππβ¦ π½ππ = πππππππ ππ πππ π π· = π × ππ + π = πππ π½ 5. In the following circuit find its Thevenin and Norton equivalent 2 4V Ans. DC π½ππ = π + π × π = ππ½ 3 2A πΉππ = π + π = πβ¦ 3 2 2A DC 2 Isc 3 RTH Short circuit current is due to voltage source (4 Volts) and current source(2 Amp.) Using superposition π Short circuit current due to voltage source = π π×π π Short circuit current due to current source = π+π = π π π π π°ππ = π + π = π πππ Thevenin’s and Norton’s equivalents are shown below: Rth VTH Isc DC Thevenin s equivalent Rth Norton s equivalent 6. For the following circuit find the value of vTH and RTH i1 6 3i1 + vTH, RTH 4 - Ans. Circuit does not have any independent source, So, π½ππ = π. For finding πΉππ connect test voltage source Vtest at terminals XY supplying 1A ππ = −ππ¨ 6 3i1 P + i1 4 - DC 1A Writing node equation at node P π½ππππ π + π½ππππ −πππ π + ππ = π π π½ππππ = −π ππ = π π π½ππππ = π πΉππ = π½ππππ π π = π = π. πβ¦ Vtest 7. In the circuit shown in following figure find the value vTH and RTH -+ 4V + 0.1v1 vTH, RTH 5 v1 Ans. By source transformation - + 5v1 5 -+ 4V X v1 Y π½ππ = ππ π. π ππ + π = ππ π. π ππ − ππ = −π π ππ = π.π = π π½ππ = π πππππ πΉππ = π½ππ π°πΊπͺ , πππ is short circuit current. By putting short across πΏπ, ππ = π, so π π , πππ = π ππ πΉππ = π½ππ πππ = π π⁄ π = ππβ¦ dependent voltage source π. π ππ = 8. In the following circuit find the effective resistance faced by the voltage source 4 i vS DC π= Ans. i/4 π½πΊ π π +π ππ = π½πΊ + π ππ = π½πΊ π= π½πΊ π The effective resistance faced by the voltage source is 3β¦ 9. Find the value of R (in ohms) for maximum power transfer in the network shown in the figure. 4 5 20 3A R 25V Ans. For maximum power transfer πΉπ³ = πΉπΊ (resistance looking into the network) Replacing independent sources by their internal impedances. 5 4 20 R πΉπΊ = πβππ + π =π+π = πβ¦ πΉ = πβ¦ 10. Find the Thevenin equivalent impedance ZTh between the nodes P and Q in the following circuit 1F 1H P 1 1 DC 1A Q 10V Ans. For finding πππ deactivate all the active sources 1/S S P 1 1 Q 1A πππ (Between P & Q nodes) π = (πΊ + π) β(π + πΊ) = πβ¦ 11. Find the value of RL such that the power transferred to RL is maximum. 10 10 10 DC RL 1A DC For maximum power transfer πΉπ³ = πΉππ Ans. 10 10 10 1A Rth πΉππ = ππβππ + ππ = π + ππ = ππβ¦ πΉπ³ = ππβ¦ 12. A source vs(t)=Vcos100 πt has an internal impedance of (4+j3)β¦. If an purely resistive load connected to this source has to extract the maximum power out of the source, find it’s value Ans. For pure resistive load RL to extract the maximum power πΉπ³ = √πΉππΊ + πΏππΊ = √ππ + ππ = πβ¦
