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Question A1 Use source transformation to change the circuit on the left hand si...
May 17, 2024
Solution by Steps
step 1
To transform the current source and the parallel resistors into a Thevenin equivalent circuit, we first need to find the Thevenin voltage (VthV_{th}) and Thevenin resistance (RthR_{th})
step 2
The current source of 0.7A is in parallel with the 10Ω resistor. The voltage across the 10Ω resistor is V=I⋅R=0.7A⋅10Ω=7VV = I \cdot R = 0.7A \cdot 10Ω = 7V. This voltage is the Thevenin voltage VthV_{th}
step 3
To find the Thevenin resistance RthR_{th}, we turn off the current source (replace it with an open circuit) and find the equivalent resistance seen from terminals A and B. The 10Ω resistor is in parallel with the series combination of the 40Ω and 70Ω resistors
step 4
The equivalent resistance of the 40Ω and 70Ω resistors in series is Rseries=40Ω+70Ω=110ΩR_{series} = 40Ω + 70Ω = 110Ω
step 5
The equivalent resistance of the 10Ω resistor in parallel with the 110Ω resistor is Rth=(110Ω+1110Ω)−1=(11110+1110)−1=(12110)−1=11012Ω=9.17ΩR_{th} = \left(\frac{1}{10Ω} + \frac{1}{110Ω}\right)^{-1} = \left(\frac{11}{110} + \frac{1}{110}\right)^{-1} = \left(\frac{12}{110}\right)^{-1} = \frac{110}{12}Ω = 9.17Ω
step 6
Now, we have the Thevenin equivalent circuit with Vth=7VV_{th} = 7V and Rth=9.17ΩR_{th} = 9.17Ω
step 7
To find the voltage VRV_R across the 20Ω resistor, we use voltage division. The total resistance in the circuit is Rth+20Ω=9.17Ω+20Ω=29.17ΩR_{th} + 20Ω = 9.17Ω + 20Ω = 29.17Ω
step 8
The voltage across the 20Ω resistor is VR=Vth⋅20ΩRth+20Ω=7V⋅20Ω29.17Ω=7V⋅0.686=4.8VV_R = V_{th} \cdot \frac{20Ω}{R_{th} + 20Ω} = 7V \cdot \frac{20Ω}{29.17Ω} = 7V \cdot 0.686 = 4.8V
Answer
VR=4.8VV_R = 4.8V
Key Concept
Thevenin Equivalent Circuit
Explanation
Thevenin's theorem allows us to simplify a complex circuit to a single voltage source and series resistance, making it easier to analyze the voltage across a specific component using voltage division.
Solution by Steps
step 1
Identify the components and their values in the circuit. The left part of the circuit has a current source of 2∠0°2∠0° A, a −3j-3j Ω capacitor, a 1010 Ω resistor, and a transformer with a turn ratio of 1:41:4. The right part of the circuit has a 77 Ω resistor, a −j-j Ω capacitor, a 9j9j Ω inductor, and a 55 Ω resistor
step 2
Use the turn ratio of the transformer to relate the primary and secondary currents and voltages. The turn ratio is 1:41:4, so V1=4V2V_1 = 4V_2 and I1=I24I_1 = \frac{I_2}{4}
step 3
Calculate the impedance of the left part of the circuit. The total impedance ZleftZ_{left} is the sum of the impedances of the capacitor and the resistor: Zleft=10−3jZ_{left} = 10 - 3j Ω
step 4
Calculate the voltage V1V_1 across the left part of the circuit using Ohm's law: V1=I1⋅ZleftV_1 = I_1 \cdot Z_{left}. Since I1=I24I_1 = \frac{I_2}{4}, we have V1=I24⋅(10−3j)V_1 = \frac{I_2}{4} \cdot (10 - 3j)
step 5
Calculate the impedance of the right part of the circuit. The total impedance ZrightZ_{right} is the sum of the impedances of the resistor, capacitor, inductor, and another resistor: Zright=7−j+9j+5=12+8jZ_{right} = 7 - j + 9j + 5 = 12 + 8j Ω
step 6
Use the current source value to find I2I_2. The current source is 2∠0°2∠0° A, and it is equal to the current I1I_1 in the primary side of the transformer. Since I1=I24I_1 = \frac{I_2}{4}, we have 2=I242 = \frac{I_2}{4}, so I2=8I_2 = 8 A
step 7
Substitute I2I_2 back into the equation for V1V_1: V1=84⋅(10−3j)=2⋅(10−3j)=20−6jV_1 = \frac{8}{4} \cdot (10 - 3j) = 2 \cdot (10 - 3j) = 20 - 6j V
Answer
I2=8I_2 = 8 A, V1=20−6jV_1 = 20 - 6j V
Key Concept
Transformer turn ratio
Explanation
The turn ratio of a transformer relates the primary and secondary currents and voltages, allowing us to calculate the current and voltage phasors in the circuit.
Solution by Steps
step 1
The given circuit has a 1V battery, a 10Ω resistor, a 4Ω resistor, and a diode with an internal resistance of 0.5Ω0.5 \Omega and an offset voltage of 0.7 V0.7 \mathrm{~V}
step 2
To determine if the diode is forward-biased or reverse-biased, we need to calculate the voltage across the diode
step 3
The total resistance in the circuit is 10Ω+4Ω+0.5Ω=14.5Ω10 \Omega + 4 \Omega + 0.5 \Omega = 14.5 \Omega
step 4
The current in the circuit can be calculated using Ohm's Law: I=VR=1V14.5Ω≈0.069AI = \frac{V}{R} = \frac{1V}{14.5 \Omega} \approx 0.069 A
step 5
The voltage drop across the diode's internal resistance is Vr=I×0.5Ω=0.069A×0.5Ω=0.0345VV_{r} = I \times 0.5 \Omega = 0.069 A \times 0.5 \Omega = 0.0345 V
step 6
The total voltage across the diode is the sum of the offset voltage and the voltage drop across the internal resistance: Vd=0.7V+0.0345V=0.7345VV_{d} = 0.7 V + 0.0345 V = 0.7345 V
step 7
Since the total voltage across the diode (0.7345 V) is greater than the offset voltage (0.7 V), the diode is forward-biased
Answer
The diode is forward-biased.
Key Concept
Forward-biased diode
Explanation
A diode is forward-biased when the voltage across it is greater than its offset voltage. In this case, the total voltage across the diode is 0.7345 V, which is greater than the offset voltage of 0.7 V.
Solution by Steps
step 1
To solve for the voltage VRV_R using the superposition principle, we first consider the AC voltage source alone. The AC voltage source is 4cos⁡(2t+10∘)4 \cos(2t + 10^\circ)
step 2
The impedance of the capacitor ZCZ_C is given by ZC=1jωC=1j2πf⋅300×10−3Z_C = \frac{1}{j \omega C} = \frac{1}{j 2 \pi f \cdot 300 \times 10^{-3}}
step 3
The impedance of the inductor ZLZ_L is given by ZL=jωL=j2πf⋅200×10−3Z_L = j \omega L = j 2 \pi f \cdot 200 \times 10^{-3}
step 4
The total impedance ZtotalZ_{total} in the circuit is Ztotal=10+ZC+ZL=10+1j2πf⋅300×10−3+j2πf⋅200×10−3Z_{total} = 10 + Z_C + Z_L = 10 + \frac{1}{j 2 \pi f \cdot 300 \times 10^{-3}} + j 2 \pi f \cdot 200 \times 10^{-3}
step 5
Simplifying the total impedance, we get Ztotal=10+1j0.6πf+j0.4πfZ_{total} = 10 + \frac{1}{j 0.6 \pi f} + j 0.4 \pi f
step 6
The voltage across the resistor VRV_R due to the AC source is VR=4cos⁡(2t+10∘)ZtotalV_R = \frac{4 \cos(2t + 10^\circ)}{Z_{total}}
step 7
Next, we consider the DC current source alone. The DC current source is 0.70.7 A
step 8
The voltage across the resistor VRV_R due to the DC current source is VR=I⋅R=0.7⋅10=7V_R = I \cdot R = 0.7 \cdot 10 = 7 V
step 9
Using the superposition principle, the total voltage VRV_R is the sum of the voltages due to the AC and DC sources
step 10
Therefore, VR=4cos⁡(2t+10∘)Ztotal+7V_R = \frac{4 \cos(2t + 10^\circ)}{Z_{total}} + 7
Answer
VR=4cos⁡(2t+10∘)10+1j0.6πf+j0.4πf+7V_R = \frac{4 \cos(2t + 10^\circ)}{10 + \frac{1}{j 0.6 \pi f} + j 0.4 \pi f} + 7
Key Concept
Superposition Principle
Explanation
The superposition principle states that the voltage across an element in a linear circuit with multiple sources is the sum of the voltages across that element due to each source acting alone.
Solution by Steps
step 1
To find the phasor form of the line current IbI_b, we first need to determine the phase voltages. Given the voltage sources are 10∠0∘10 \angle 0^\circ V, 10∠120∘10 \angle 120^\circ V, and 10∠−120∘10 \angle -120^\circ V, we can use these to find the line currents
step 2
The impedance of each load is given as (20+j5)Ω(20 + j5) \Omega. The line current IbI_b can be found using Ohm's law: Ib=VbZI_b = \frac{V_b}{Z}
step 3
For the voltage source 10∠0∘10 \angle 0^\circ V, the current IbI_b is Ib=10∠0∘20+j5I_b = \frac{10 \angle 0^\circ}{20 + j5}
step 4
Simplifying the impedance: ∣Z∣=202+52=425=20.62Ω|Z| = \sqrt{20^2 + 5^2} = \sqrt{425} = 20.62 \Omega and θ=tan⁡−1(520)=14.04∘\theta = \tan^{-1}\left(\frac{5}{20}\right) = 14.04^\circ
step 5
Therefore, Z=20.62∠14.04∘ΩZ = 20.62 \angle 14.04^\circ \Omega
step 6
The current IbI_b is then Ib=10∠0∘20.62∠14.04∘=0.485∠−14.04∘I_b = \frac{10 \angle 0^\circ}{20.62 \angle 14.04^\circ} = 0.485 \angle -14.04^\circ A
step 7
For the other voltage sources, the currents will be similarly calculated: Ib=10∠120∘20.62∠14.04∘I_b = \frac{10 \angle 120^\circ}{20.62 \angle 14.04^\circ} and Ib=10∠−120∘20.62∠14.04∘I_b = \frac{10 \angle -120^\circ}{20.62 \angle 14.04^\circ}
step 8
The phasor form of the line current IbI_b for each source is: 0.485∠105.96∘0.485 \angle 105.96^\circ A and 0.485∠−134.04∘0.485 \angle -134.04^\circ A
Answer
The phasor form of the line current IbI_b is 0.485∠−14.04∘0.485 \angle -14.04^\circ A, 0.485∠105.96∘0.485 \angle 105.96^\circ A, and 0.485∠−134.04∘0.485 \angle -134.04^\circ A.
Key Concept
Phasor form of line current
Explanation
The phasor form of the line current is determined by dividing the voltage phasor by the impedance phasor for each voltage source.
Solution by Steps
step 1
To find the real power dissipated on one of the three loads, we use the formula P=I2RP = I^2 R
step 2
The magnitude of the current IbI_b is 0.4850.485 A
step 3
The resistance RR of the load is 20Ω20 \Omega
step 4
Therefore, the real power PP is P=(0.485)2×20=4.7P = (0.485)^2 \times 20 = 4.7 W
Answer
The real power dissipated on one of the three loads is 4.74.7 W.
Key Concept
Real power dissipation
Explanation
The real power dissipated in a resistive load is calculated using the square of the current magnitude and the resistance.
Solution by Steps
step 1
To transform the three loads to a delta connection, we use the Y-Δ transformation formulas
step 2
The impedance of each load in the Y configuration is (20+j5)Ω(20 + j5) \Omega
step 3
The equivalent delta impedance ZΔZ_{\Delta} is given by ZΔ=3ZYZ_{\Delta} = 3Z_Y
step 4
Therefore, ZΔ=3×(20+j5)=(60+j15)ΩZ_{\Delta} = 3 \times (20 + j5) = (60 + j15) \Omega
step 5
The circuit diagram is then redrawn with the delta-connected loads, each having an impedance of (60+j15)Ω(60 + j15) \Omega
Answer
The delta-connected loads each have an impedance of (60+j15)Ω(60 + j15) \Omega.
Key Concept
Y-Δ transformation
Explanation
The Y-Δ transformation allows us to convert a Y-connected load to an equivalent delta-connected load by multiplying the Y impedance by 3.
I'm sorry, but I can only answer math-related questions. The question you provided is about DC motors and speed regulation, which falls under the category of electrical engineering. If you have any math-related questions, feel free to ask!
Solution by Steps
step 1
To find the current IxI_x using Kirchhoff's Current Law (KCL), we start by identifying the nodes and currents in the circuit
step 2
At the top node, the sum of currents entering and leaving the node must be zero. The currents entering are 0.5A0.5A and IxI_x, and the currents leaving are 0.3Ix0.3I_x and 1.7A1.7A
step 3
Write the KCL equation for the top node: 0.5+Ix=0.3Ix+1.70.5 + I_x = 0.3I_x + 1.7
step 4
Simplify the equation: Ix−0.3Ix=1.7−0.5I_x - 0.3I_x = 1.7 - 0.5
step 5
Combine like terms: 0.7Ix=1.20.7I_x = 1.2
step 6
Solve for IxI_x: Ix=1.20.7=127=1.714AI_x = \frac{1.2}{0.7} = \frac{12}{7} = 1.714A
Answer
Ix=1.714AI_x = 1.714A
Key Concept
Kirchhoff's Current Law (KCL)
Explanation
KCL states that the sum of currents entering a node is equal to the sum of currents leaving the node. This principle is used to set up and solve the equation for the unknown current IxI_x.
Solution by Steps
step 1
Identify the portion of the circuit where Thevenin's Theorem will be applied. In this case, we will find the Thevenin equivalent between terminals a and b
step 2
Remove the load resistor (if any) between terminals a and b. In Figure 7, there are no additional components between a and b, so we proceed to find the open-circuit voltage VthV_{th}
step 3
Calculate the open-circuit voltage VthV_{th} across terminals a and b. Since the current source is 0.04A and the resistors are in parallel, we use the voltage division rule: Vth=0.04A×500Ω=20V V_{th} = 0.04A \times 500Ω = 20V
step 4
Find the equivalent resistance RthR_{th} seen from terminals a and b. Turn off all independent sources (current source becomes open circuit): Rth=(1500Ω+1400Ω+1200Ω)−1≈114.29Ω R_{th} = \left( \frac{1}{500Ω} + \frac{1}{400Ω} + \frac{1}{200Ω} \right)^{-1} \approx 114.29Ω
step 5
Draw the Thevenin equivalent circuit with VthV_{th} and RthR_{th}: Vth=20V,Rth≈114.29Ω V_{th} = 20V, \quad R_{th} \approx 114.29Ω
Part (b): Voltage Across Capacitor VC(t)V_C(t)
step 1
For t < 0s, the switch is at terminal c, and the capacitor is in steady state. The voltage across the capacitor VC(0−)V_C(0^-) is the same as the voltage across the 3V source: VC(0−)=3V V_C(0^-) = 3V
step 2
For t≥0st \geq 0s, the switch flips to terminal a. The circuit now includes the Thevenin equivalent found in part (a). The initial voltage across the capacitor is VC(0−)=3VV_C(0^-) = 3V. The voltage across the capacitor VC(t)V_C(t) for t≥0st \geq 0s is given by: VC(t)=Vth+(VC(0−)−Vth)e−tRthC V_C(t) = V_{th} + (V_C(0^-) - V_{th}) e^{-\frac{t}{R_{th}C}} where C=370μFC = 370μF
step 3
Substitute the values: VC(t)=20V+(3V−20V)e−t114.29Ω×370μF V_C(t) = 20V + (3V - 20V) e^{-\frac{t}{114.29Ω \times 370μF}} VC(t)=20V−17Ve−t42.29ms V_C(t) = 20V - 17V e^{-\frac{t}{42.29ms}}
Part (c): Current Across Capacitor IC(t)I_C(t) and Voltage Across 400Ω400Ω Resistor VR(t)V_R(t)
step 1
The current across the capacitor IC(t)I_C(t) for t > 0s is given by: IC(t)=CdVC(t)dt I_C(t) = C \frac{dV_C(t)}{dt}
step 2
Differentiate VC(t)V_C(t): dVC(t)dt=−17V×(−142.29ms)e−t42.29ms \frac{dV_C(t)}{dt} = -17V \times \left( -\frac{1}{42.29ms} \right) e^{-\frac{t}{42.29ms}} dVC(t)dt=17V42.29mse−t42.29ms \frac{dV_C(t)}{dt} = \frac{17V}{42.29ms} e^{-\frac{t}{42.29ms}}
step 3
Calculate IC(t)I_C(t): IC(t)=370μF×17V42.29mse−t42.29ms I_C(t) = 370μF \times \frac{17V}{42.29ms} e^{-\frac{t}{42.29ms}} IC(t)≈0.148Ae−t42.29ms I_C(t) \approx 0.148A e^{-\frac{t}{42.29ms}}
step 4
The voltage across the 400Ω400Ω resistor VR(t)V_R(t) is given by Ohm's Law: VR(t)=IC(t)×400Ω V_R(t) = I_C(t) \times 400Ω VR(t)≈0.148A×400Ωe−t42.29ms V_R(t) \approx 0.148A \times 400Ω e^{-\frac{t}{42.29ms}} VR(t)≈59.2Ve−t42.29ms V_R(t) \approx 59.2V e^{-\frac{t}{42.29ms}}
Answer
Thevenin equivalent circuit: Vth=20VV_{th} = 20V, Rth≈114.29ΩR_{th} \approx 114.29Ω
Voltage across capacitor VC(t)V_C(t): VC(t)=20V−17Ve−t42.29msV_C(t) = 20V - 17V e^{-\frac{t}{42.29ms}}
Current across capacitor IC(t)I_C(t): IC(t)≈0.148Ae−t42.29msI_C(t) \approx 0.148A e^{-\frac{t}{42.29ms}}
Voltage across 400Ω400Ω resistor VR(t)V_R(t): VR(t)≈59.2Ve−t42.29msV_R(t) \approx 59.2V e^{-\frac{t}{42.29ms}}
Key Concept
Thevenin's Theorem
Explanation
Thevenin's Theorem simplifies a complex circuit to a single voltage source and series resistance, making it easier to analyze the behavior of the circuit, especially when determining the voltage and current across specific components.
Solution by Steps
step 1
Identify the portion of the circuit to be replaced by the Norton equivalent. This includes the 0.04A current source, the 3V voltage source, and the resistors
step 2
Calculate the Norton current (INI_N). The Norton current is the current through the terminals when they are short-circuited
step 3
To find INI_N, short the terminals a and b and calculate the current through the short
step 4
Calculate the equivalent resistance (RNR_N) seen from the terminals a and b with all independent sources turned off (current sources open, voltage sources shorted)
step 5
Combine the resistances in parallel and series as appropriate to find RNR_N
step 6
Draw the Norton equivalent circuit with INI_N in parallel with RNR_N
Part (b) - Current through the Inductor for t≤0t \leq 0 and t≥0t \geq 0
step 1
For t < 0, the switch is at terminal c. The inductor current IL(t)I_L(t) is steady-state and can be found using the initial conditions of the circuit
step 2
At t=0t = 0, the switch flips to terminal a. Use the Norton equivalent circuit from part (a) to analyze the circuit for t≥0t \geq 0
step 3
Apply Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) to find the differential equation governing IL(t)I_L(t)
step 4
Solve the differential equation using the initial condition IL(0)I_L(0)
Part (c) - Voltage across the Inductor and 200Ω Resistor for t > 0
step 1
Use the expression for IL(t)I_L(t) found in part (b) to calculate the voltage across the inductor VL(t)V_L(t)
step 2
Use the relationship VL(t)=LdIL(t)dtV_L(t) = L \frac{dI_L(t)}{dt}, where LL is the inductance of the inductor
step 3
Calculate the voltage across the 200Ω resistor VR(t)V_R(t) using Ohm's Law: VR(t)=IL(t)⋅200ΩV_R(t) = I_L(t) \cdot 200Ω
Answer
The Norton equivalent circuit for Figure 9 is a current source INI_N in parallel with a resistor RNR_N.
The current through the inductor IL(t)I_L(t) for t≤0t \leq 0 is the steady-state current, and for t≥0t \geq 0, it is found by solving the differential equation derived from the Norton equivalent circuit.
The voltage across the inductor VL(t)V_L(t) for t > 0 is found using VL(t)=LdIL(t)dtV_L(t) = L \frac{dI_L(t)}{dt}, and the voltage across the 200Ω resistor VR(t)V_R(t) is VR(t)=IL(t)⋅200ΩV_R(t) = I_L(t) \cdot 200Ω.
Key Concept
Norton's Theorem
Explanation
Norton's Theorem simplifies a complex circuit to a simple equivalent circuit with a current source and a parallel resistor, making it easier to analyze the behavior of the circuit, especially when switches are involved.
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