
Concept explainers
(a)
Find the two-component Thevenin equivalent of the network
(a)

Answer to Problem 63E
The Thevenin equivalent resistance is
Explanation of Solution
Formula used:
The expression for the equivalent resistor when resistors are connected in series is as follows:
Here,
The expression for the equivalent resistor when resistors are connected in parallel is as follows:
Here,
The
o
Refer to the redrawn Figure 1:
The expression for the conversion of
Here,
Calculation:
To find equivalent resistance of a circuit the independent voltage source is replaced by short circuit
The redrawn circuit diagram is given in Figure 2:
Refer to the redrawn Figure 2:
Substitute
Rearrange the equation for
The simplified circuit diagram is given in Figure 3:
Refer to the redrawn Figure 3:
Substitute
Substitute
Substitute
The simplified circuit diagram is given in Figure 4.
Refer to the redrawn Figure 4:
Substitute
Substitute
Substitute
Rearrange the equation for
The simplified circuit diagram is given in Figure 5.
Refer to the redrawn Figure 5:
Substitute
The simplified circuit diagram is given in Figure 6.
Refer to the redrawn Figure 6:
So, the Thevenin equivalent resistance is
The redrawn circuit diagram is given in Figure 7.
Refer to the redrawn Figure 7:
Apply KCL at node 1:
Here,
Substitute
Rearrange for
Apply KCL at node 2:
Here,
Substitute
Rearrange for
Apply KCL at node 3:
Here,
Substitute
Rearrange for
The equations (7), (9) and (11) can be written in matrix form as:
Therefore, by Cramer’s rule,
The determinant of the coefficient matrix is as follows:
The 1st determinant is as follows:
The 2nd determinant is as follows:
The 3rd determinant is as follows:
Simplify for
Simplify for
Simplify for
So, the Thevenin voltage
Conclusion:
Thus, the Thevenin equivalent resistance is
(b)
Find the power dissipated by a
(b)

Answer to Problem 63E
Thepower dissipated by a
Explanation of Solution
Given Data:
The load resistance is
Formula used:
The expression for the power dissipated by a resistor is as follows:
Here,
Calculation:
The redrawn circuit diagram is given in Figure 8.
Refer to the redrawn Figure 8:
The expression for the current flowing in the circuit is as follows:
Here,
Substitute
Substitute
Conclusion:
Thus, the power dissipated by a
Want to see more full solutions like this?
Chapter 5 Solutions
ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<
- An FDM is used to multiplex two groups of signals using AM-SSB, the first group contains 25 speech signals, each has maximum frequency of 4 kHz, the second group contains 15 music signals, each has maximum frequency of 10 kHz. A guard bandwidth of 500 Hz is used between each two signals and before the first one. 1. Find the BWmultiplexing 2. Find the BWtransmission if the multiplexing signal is modulated using AM-DSB-LC.arrow_forwardA single tone is modulated using FM transmitter. The SNR; at the input of the demodulator Is 20 dB. If the maximum frequency of the modulating signal is 4 kHz, and the maximum frequency deviation is 12 kHz, find the SNR, and the bandwidth (using Carson rule) at the following conditions: 1. For the given values of fm and Af. 2. If the amplitude of the modulating signal is increased by 80%. 3. If the amplitude of the modulating signal is decreased by 50%, and frequency of modulating signal is increased by 50%.arrow_forwardFM station of 100 MHz carrier frequency modulated by a 20 kHz sinusoid with an amplitude of 10 volt, so that the peak frequency deviation is 25 kHz determine: 1) The BW of the FM signal. 2) The approximated BW if the modulating signal amplitude is increased to 50 volt. 3) The approximated BW if the modulating signal frequency is increased by 70%. 4) The amplitude of the modulating signal if the BW is 65 kHz.arrow_forward
- b) The joint probability function for the random variables X and Y is given in Table below. Find a) the marginal probability function of X and Y. P(Y/X) and P(X/Y). c) P(X ≥ 2, Y ≤ 2) y 1 2 3 10.05 0.05 0.1 P(X, Y) = X 20.05 0.1 0.35 3 0 0.2 0.1arrow_forwardSuppose a random variable X as pmf / Px (x) = { %, x = 1, 2, 3, 0, otherwise. find constand c ①P(X = 1), P(X 7,2), PC1 3) C CDFarrow_forwardSuppose that a coin is tossed three so that the sample space is Let X represent the number of heads that can come up. i) Find the probability function corresponding to the random variable X. Assuming that the coin is fair ii) Find the distribution function for the random variable X. iii) Obtain its graph.arrow_forward
- Q9 A single-phase transformer, 2500 / 250 V, 50 kVA, 50 Hz has the following parameters, the Primary and secondary resistances are 0.8 ohm and 0.012 ohm respectively, the primary and secondary reactance are 4 ohm and 0.04 ohm respectively and the transformer gives 96% maximum efficiency at 75% full-load. The magnetizing component of-load current is 1.2 A on 2500 V side. 1- Draw the equivalent circuit referred to primary (H.V side) and inserts all the values in it 2- Find out Ammeter, voltmeter and wattmeter readings on open-circuit and short-circuit test. If supply is given to 2500 V side in both cases. Ans. O.C. Test (Vo= 2500 V, lo=1.24 A, Wo=781.25 w) S.C. Test (Vsc =164.924 V, Isc =20 A, Wsc =800 w )arrow_forwardQ2-A)- Enumerate the various losses in transformer. Explain how each loss varies with (Load current, supply voltage). B)- Draw the pharos diagram at load on primary side.arrow_forwardQ2- What are the parameters and loss that can be determined during open-circuit test of singlephase transformer. Draw the circuit diagram of open-circuit test and explain how can you calculate the Parameters and loss.arrow_forward
- Q2-Drive the condition of maximum efficiency of single-phase transformer. Q1- A 5 KVA, 500/250 V ,50 Hz, single phase transformer gave the following reading: O.C. Test: 250 V,2 A, 50 W (H.V. side open) S.C. Test: 25 V10 A, 60 W (L.V. side shorted) Determine: i) The efficiency on full load, 0.8 lagging p.f. ii) Draw the equivalent circuit referred to primary and insert all the values it.arrow_forwardQ2- Describe various losses in transformer. Explain how each loss varies with load current, supply voltagearrow_forwardQ1-A 12 KVA, 440/ 220 V, 50 Hz single phase transformer has 275 secondary turns. The no load current of transformer is 2A at power factor 0.375 when connected to 220 V, 50 Hz supply. The full load copper loss is 198.3 watt. Calculate a) Maximum value of flux in the core. b) Maximum efficiency at 0.8 lagging p.f c) KVA supply at maximum efficiencyarrow_forward
- Introductory Circuit Analysis (13th Edition)Electrical EngineeringISBN:9780133923605Author:Robert L. BoylestadPublisher:PEARSONDelmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage LearningProgrammable Logic ControllersElectrical EngineeringISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education
- Fundamentals of Electric CircuitsElectrical EngineeringISBN:9780078028229Author:Charles K Alexander, Matthew SadikuPublisher:McGraw-Hill EducationElectric Circuits. (11th Edition)Electrical EngineeringISBN:9780134746968Author:James W. Nilsson, Susan RiedelPublisher:PEARSONEngineering ElectromagneticsElectrical EngineeringISBN:9780078028151Author:Hayt, William H. (william Hart), Jr, BUCK, John A.Publisher:Mcgraw-hill Education,





