To determine Calculate the total power developed in the given circuit. Explanation Given data: Refer to Figure P2.9 in the textbook for required data. v o = 5 V Formula used: Write the expression for power developed or absorbed by the source (voltage or current) as follows: p = v s i s (1) Here, i s is the current through the source and v s is the voltage across the source. Calculation: From the given circuit, voltage of dependent source v o is equal to 10 v a . Therefore, write the expression as follows: 10 v a = v o v a = v o 10 Substitute 5 V for v o to obtain the value of v a . v a = 5 V 10 = 0.5 V As all the branches of the circuit are in parallel, the voltage across each branch is same. Therefore, write the expression to find the voltage drop across 9 A current source as follows: v 9 A + 20 V = 5 V Simplify the expression to find the voltage drop across 9 A current source as follows: v 9 A = 5 V − 20 V = − 15 V From the circuit, write the expression to find the voltage of v g voltage source as follows: − v g + v a = 5 V Substitute 0.5 V or v a and simplify the expression as follows: − v g + 0.5 V = 5 V v g = − 4.5 V From the calculations, redraw the circuit as shown in Figure 1. If the current enters from the negative terminal and leaves from the positive terminal of a voltage source then the source delivers the power. The delivering power is obtained with negative sign. If the current enters from the positive terminal and leaves from the negative terminal of a voltage source then the source absorbs the power. The absorbing power is obtained with positive sign. Rewrite the expression in Equation (1) to find the power associated with the 9 A current source as follows: p 9 A = ( v s ) 9 A ( i s ) 9 A Substitute − 15 V for ( v s ) 9 A and 9 A for ( i s ) 9 A to find the power associated with the 9 A current source as follows. p 9 A = ( − 15 V ) ( 9 A ) = − 135 VA = − 135 W { ∵ 1 VA=1 W } As the power is obtained with negative sign, the power is developed power. Therefore, the power developed by the 9 A current source is 135 W. p 9 A = 135 W Rewrite the expression in Equation (1) to find the power associated with the 20 V voltage source as follows: p 20 V = ( v s ) 20 V ( i s ) 20 V Substitute 20 V for ( v s ) 20 V and 9 A for ( i s ) 20 V to find the power associated with the 20 V voltage source as follows. p 20 V = ( 20 V ) ( 9 A ) = 180 VA = 180 W As the power is obtained with positive sign, the power is absorbed power

ELECTRIC CIRCUITS-W/MASTERINGENGINEERING

11th Edition

ISBN: 9780134894300

Author: NILSSON

Publisher: PEARSON

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Chapter 2, Problem 9P

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ELECTRIC CIRCUITS-W/MASTERINGENGINEERING

Ch. 2.1 - Prob. 1AP Ch. 2.1 - For the circuit shown, What value of α is required...Ch. 2.2 - For the circuit shown, If υg = 1 kV and ig = 5 mA,...Ch. 2.2 - For the circuit shown, If ig = 0.5 A and G = 50...Ch. 2.4 - Prob. 5AP Ch. 2.4 - Use Ohm’s law and Kirchhoff’s laws to find the...Ch. 2.4 - a) The terminal voltage and terminal current were...Ch. 2.4 - Repeat Assessment Problem 2.7, but use the...Ch. 2.5 - Prob. 9AP Ch. 2.5 - The current iϕ in the circuit shown is 2 A....

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Kirchhoff's Rules of Electrical Circuits; Author: Flipping Physics;https://www.youtube.com/watch?v=d0O-KUKP4nM;License: Standard YouTube License, CC-BY