Explanation Formula used: The expression for parallel combination of resistance is as follows. 1 R = ( 1 R 1 + 1 R 2 ) (1) Here, R is the total resistances and R 1 and R 2 are the resistances in the circuit. The expression for voltage is as follows. v = i R (2) Here, v is the voltage , i is the current and R is value of resistance. The expression for resistance is as follows. R = v i (3) Here, v is magnitude of the voltage, i is the magnitude of current and R is value of resistance. Calculation: Refer to FIGURE 5.85 in the textbook, Thevenin resistance for the circuit is resistor of value 2 MΩ in parallel with equivalent resistance of remaining circuit. First remove 2 MΩ resistor and as circuit contains dependent current source, connect a current source of 1 A value across open circuited terminal to find equivalent resistance for remaining network. The circuit diagram is redrawn as shown in Figure 1, Refer to the redrawn Figure 1, The expression for voltage v π is as follows, v π = i 2 R 2 (4) Here, v π is the voltage across resistor R 2 , i 2 is the current in the circuit and R 2 is value of resistance in the circuit. Substitute 1 A for i 2 and r π for R 2 in equation (4), v π = ( 1 A ) ( r π ) = r π V The function for current i 1 is as follows, i 1 = 0.02 v π (5) Here, i 1 is the current of dependent source and v π is the voltage across resistor R 2 . Substitute r π V for v π in equation (5), i 1 = 0.02 ( r π V ) = 0.02 r π A As resistor R 3 and resistor R 4 are in parallel therefore from the formula in equation (1), 1 R 5 = ( 1 R 3 + 1 R 4 ) 1 R 5 = 1 1 kΩ + 1 2 kΩ 1 R 5 = 1 0.667 kΩ (6) Rearrange equation (6) for R 5 , R 5 = 667 Ω The circuit diagram is redrawn as shown in Figure 2, Refer to Figure 2, Substitute 0.02 r π A for i and 667 Ω for R in equation (2), v = ( 0.02 r π A ) 667 Ω = 13.34 r π V So value of voltage v 1 is 13.34 r π V . Current source in parallel with a resistor can change to a voltage source connected with same resistor in series. The circuit diagram is redrawn as shown in Figure 3, Refer to the redrawn Figure 3, The expression for KVL in mesh D A B C D is as follows, − v + i 2 R 2 + i 2 R 5 + v 1 = 0 (7) Here, v 1 is the voltage in loop D A B C D , v is voltage across current source i 2 , i 2 is the current flowing in loop D A B C D and R 2 and R 5 are the resistances in the circuit

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ISBN: 9781260540666

Author: Hayt

Publisher: MCG CUSTOM

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The Z-parameters that are the impedance parameters of a two-port network are defined by expressing the two-port voltages V1 and V2 in terms of two-port currents I1 and I2. The Z-parameters are calculated by open circuiting one of the two ports at a time …

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Chapter 5, Problem 44E

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Find the value of Thevenin equivalent resistance of the circuit to the right of the dashed line.

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Chapter 5, Problem 43E

Chapter 5, Problem 45E

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ENGINEERING CIRCUIT...(LL)>CUSTOM PKG.<

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