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Concept explainers
In the circuit of Fig. 7.144, find the value of io for all values of 0 < t.
![Check Mark](/static/check-mark.png)
Find the current
Answer to Problem 80P
The current
Explanation of Solution
Given data:
Refer to Figure 7.144 in the textbook.
The value of capacitance
The source voltage
The current source
Formula used:
Write the general expression to find the complete voltage response for an RC circuit.
Here,
Write the expression to find the time constant for an RC circuit.
Here,
C is the capacitance of the capacitor.
Write the general expression for the unit step function.
Calculation:
The given source voltage is,
Apply the unit step function in equation (3) to equation (4).
For
The given Figure 7.144 is redrawn as shown in Figure 1.
In Figure 1, the capacitor reaches steady state and it will acts as an open circuit. The initial voltage across the capacitor is denoted by
Apply Kirchhoff’s current law at node
Rearrange the equation as follows,
In Figure 1, the initial voltage across the capacitor
For
In Figure 2, the voltage source is equal to zero (or a short circuit). Now, the final voltage across the capacitor is represented by
Apply Kirchhoff’s current law at node
Rearrange the equation as follows,
In Figure 2, the final voltage across the capacitor
Figure 3 shows the Thevenin resistance at the capacitor terminal.
In Figure 3, the Thevenin resistance is calculated as follows.
Substitute
Substitute the units
Substitute
Figure 4 shows the modified circuit diagram.
Apply Kirchhoff’s current law at node
Substitute
Reduce the equation as follows,
Therefore, the current
Substitute
Convert the unit A to mA.
Apply the unit step function in equation (3) to equation (6).
PSpice Simulation:
For
Draw the circuit diagram in PSpice as shown in Figure 5.
Save the circuit and provide the Simulation Settings as shown in Figure 6.
Now run the simulation and the results will be displayed as shown in Figure 7 by enabling “Enable Bias Voltage Display” icon.
From Figure 7, the initial voltage across the capacitor is 17.5 V.
For
Draw the circuit diagram in PSpice as shown in Figure 8.
Now run the simulation and the results will be displayed as shown in Figure 8 by enabling “Enable Bias Voltage Display” icon.
From Figure 9, the final voltage across the capacitor is 5 V.
Draw the circuit diagram in PSpice as shown in Figure 10.
Now run the simulation and the results will be displayed as shown in Figure 11 by enabling “Enable Bias Voltage Display” icon and place the “Current Marker”
The SCHEMATIC1 dialog box is also opened with simulation result as shown in Figure 12.
Therefore, the plot of current through the
Conclusion:
Thus, the current
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Chapter 7 Solutions
Fundamentals of Electric Circuits
- A.With the aid of a diagram, describe fringing, and explain the impact that it has on the relevant magnetic circuit parameter. B. A coil of 1500 turns give rise to a magnetic flux of 2.5 mWb when carrying a certain current. If this current is reversed in 0.2 s, what is the average value of the e.m.f. induced in the coil? C.Define Mutual Inductance.Two coils are connected in series and their total inductance is measured as 0.12 H, and when the connection to one coil is reversed, the total inductance is measured as 0.04 H. If the coefficient of coupling is 0.8, determine:The self-inductance of each coil, and the mutual inductance between the coils.arrow_forwardcomparing Lenz's law and the left hand generator rule, which of these is the more important fundamental principle?arrow_forwardExample: Electric Field and Potential Inside a Charged Sphere Problem: A sphere of radius R = 0.2 m is uniformly charged with a total charge Q = 5 μC. The sphere is made of a dielectric material with relative permittivity € = 4. Calculate: 1. The electric field intensity E(r) inside and outside the sphere. 2. The electric potential (r) at any point inside the sphere. Solution: Step 1: Given Data Radius of the sphere: R = 0.2m, Total charge: Q-5 μC=5× 10° C. Step 2: Electric Field Inside the Sphere (< Using Gauss's Law:arrow_forwardplease remember to draw the circuitsarrow_forwardA balanced three-phase, A - connected induction motor consumes 3246 W when the l voltage is 208 V, and the line current is 10.6 A. Calculate: i. The motor's winding resistance. ii. The motor's winding reactance. 12 marrow_forwarda) An iron ring, having a mean circumference of 250 mm and a cross-sectional area of 400 mm², is wound with a coil of 70 turns. Using the following data, calculate the current required to set up a flux of 510µWb in the ring. H (A/m) 350 600 1250 B (T) 1.0 1.2 1.4 b) Calculate also: i. The inductance of the coil at the current obtained in Question 2 (a) above. ii. The self-induced e.m.f. if this current is switched off in 0.005 s. Assume that there is no residual flux.arrow_forwardA balanced three-phase, 1351-V, 60-Hz, A-connected source feeds a balanced Y- connected load with a per-phase impedance of 360 + j150 Q as shown in Figure 1. Calculate: i. The readings on each of the wattmeters ii. The power factor of the load using the wattmeter readings. NOTE: i. ii. Let VAN be the reference phasor, and the phase sequence be ABC anticlockwise. Assume the voltage-drop on the conductors between the source and the load to be zero volts. V b V₁ W 000 000 ; A 360 + j150 360 + j150 4 b 0000 000 B 360 + j150 C W₂ Figure 1arrow_forwarda) Three 30 2 resistors are arranged as shown in Figure 1 below. They are connected to a 480 V three-phase supply. The phase sequence is RYB anticlockwise. Calculate: i. The total power drawn by the circuit using the phase parameters. ii. The power read by each wattmeter. b) If Za, one of the 30 2 resistors, is now removed from the circuit, calculate: R- i. The line currents: IR, Iv, and la ii. The power read by each wattmeter. iii. The total power drawn by the two resistors. W₁ Be- W2 www 'R 22 12 B Figure 1arrow_forwardA certain magnetic circuit may be regarded as consisting of three parts, A, B and C in series, each one of which has a uniform cross-sectional area. Part A has a length of 300 mm and a cross- sectional area of 450 mm². Part B has a length of 120 mm and a cross-sectional area of 300 mm². Part C is an airgap 1.0 mm in length and of cross-sectional area 350 mm². The flux in the airgap is 0.35 mWb. Neglect magnetic leakage and fringing. The magnetic characteristic for parts A and B is given by: H (A/m) 400 560 800 1280 1800 B (T) 0.7 0.85 1.0 1.15 1.25 Calculate: i. The reluctance of each part, that is, of Part A, Part B, and Part C ii. The total reluctance of the magnetic circuit. iii. The total m.m.f.arrow_forwardHW_02.pdf EE 213-01 Assignments HW_#2 Toms are as muIcate uah.instructure.com b Answered: HW_#1 HW_01.pdf EE 213-01 Assignments P Pearson MyLab and Mastering uah.instructure.com P Course Home Watc... ✓ Download → Info × Close 1) (5 pts)For the circuit shown, find the value of Ia, Ib, Ic and Va: Ib 10 Ohms + Ic 40 Ohms 20 Ohms 70 Volts a Page 1 > of 2 - ZOOM + pui via Canvas Hint: use KCL to find Ic in terms of la and lb, use KVL around the right loop to find a relationship between la and lb, use KVL around the outer loop to solve for a current value (use ohms law for the voltage drops across the resistors) 2) (5 pts) – For problem 1, show that the power supplied (delivered) by the source equals the power absorbed by the resistors. 3) (5 pts) Determine the equivalent resistance looking into terminals a-b for the two circuits shown. Use series and parallel resistor combinations. Hint: work from the opposite end of terminals a-b towards terminals a-b a) 24 Ohms 10 Ohms 8 Ohms G a REQ b)…arrow_forwardA three-phase load consists of three identical impedances connected in delta. The impedances have a value of (100 + j40) Q. The supply voltage is 440 V. The two- wattmeter method is used to determine the total power drawn by the load. One wattmeter, W₁, has its current coil connected in the blue line and its voltage coil connected between the yellow and blue lines. Calculate: a) The power measured by each wattmeter b) The power factor of the load using the wattmeter readings c) The apparent power supplied to the load d) The reactive power drawn by the load.arrow_forward4- A 75 KW, 250 V shunt generator has R₁ = 0.03 2 and RF =50 2. The generator has 6 poles and is lap wound with totally 400 conductors. The generator runs at 600 rpm on full load. The bore of the machine is 42 cm (in diameter), its axial length is 28 cm and each pole subtends an angle of 54 °. Field pole a) b) c) Find the airgap flux density. (10 pt) Find the induced voltage under full load. (10 pt) Find the average number of active conductors. (5 pts) 28 cm 21 cm 60° Armaturearrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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