a) To determine Sketch the power versus time t plot for 0 ≤ t ≤ 50 s . Explanation Given data: Refer to the given plots in the problem. Calculation: The given voltage versus time t plot is drawn as Figure 1. Write the expression to calculate the straight line equation for two points ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( y − y 1 ) = y 2 − y 1 x 2 − x 1 ( x − x 1 ) (1) Refer to Figure 1, substitute t for x and v ( t ) for y in equation (1). ( v ( t ) − y 1 ) = y 2 − y 1 x 2 − x 1 ( t − x 1 ) (2) Write the expression to calculate the power. p ( t ) = v ( t ) i (3) Here, v ( t ) is the value of the voltage, and i is the value of the current. Refer to Figure 1, split up the time period as eight divisions 0 s ≤ t < 4 s , 4 s < t ≤ 8 s , 8 s ≤ t < 16 s , 16 s < t ≤ 20 s , 20 s ≤ t < 36 s , 36 s < t ≤ 46 s , 46 s ≤ t < 50 s and t > 50 s to find the respective voltage value. Case (i): 0 s ≤ t < 4 s The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 0 s , 0 V ) and ( 4 s , 1 0 V ) . Substitute 0 s for x 1 , 0 V for y 1 , 4 s for x 2 and 10 V for y 2 in equation (2). ( v ( t ) − 0 V ) = 10 V − 0 V 4 s − 0 s ( t − 0 s ) = 10 V 4 s t = 2.5 t V Simplify the above equation to find v ( t ) . v ( t ) = 2.5 t V + 0 V = 2.5 t V Case (ii): 4 s < t ≤ 8 s The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 4 s , 1 0 V ) and ( 8 s , 1 0 V ) . Substitute 4 s for x 1 , 10 V for y 1 , 8 s for x 2 and 10 V for y 2 in equation (2). ( v ( t ) − 10 V ) = 10 V − 10 V 8 s − 4 s ( t − 4 s ) = 0 V 4 s ( t − 4 s ) = 0 V Simplify the above equation to find v ( t ) . v ( t ) = 0 V + 10 V = 10 V Case (iii): 8 s ≤ t < 16 s The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 8 s , 1 0 V ) and ( 16 s , − 1 0 V ) . Substitute 8 s for x 1 , 10 V for y 1 , 16 s for x 2 and − 10 V for y 2 in equation (2). ( v ( t ) − 10 V ) = − 10 V − 10 V 16 s − 8 s ( t − 8 s ) = − 20 V 8 s ( t − 8 s ) = − 2.5 ( t − 8 s ) V = ( − 2.5 t + 20 ) V Simplify the above equation to find v ( t ) . v ( t ) = − 2.5 t V + 20 V + 10 V = − 2.5 t V + 30 V = ( − 2.5 t + 30 ) V Case (iv): 16 s < t ≤ 20 s The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 16 s , − 1 0 V ) and ( 20 s , − 1 0 V ) . Substitute 16 s for x 1 , − 10 V for y 1 , 20 s for x 2 and − 10 V for y 2 in equation (2). ( v ( t ) − ( − 10 V ) ) = ( − 10 V ) − ( − 10 V ) 20 s − 16 s ( t − 16 s ) ( v ( t ) + 10 V ) = − 10 V + 10 V 4 s ( t − 16 s ) = 0 V 4 s ( t − 16 s ) = 0 Simplify the above equation to find v ( t ) . v ( t ) = 0 − 10 V = − 10 V Case (v): 20 s ≤ t < 36 s The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 20 s , − 1 0 V ) and ( 36 s , 6 V ) . Substitute 20 s for x 1 , − 10 V for y 1 , 36 s for x 2 and 6 V for y 2 in equation (2). ( v ( t ) − ( − 10 V ) ) = 6 V − ( − 10 V ) 36 s − 20 s ( t − 20 s ) ( v ( t ) + 10 V ) = 6 V + 10 V 16 s ( t − 20 s ) = 16 V 16 s ( t − 20 s ) = 1 V ( t − 20 s ) Simplify the above equation to find v ( t ) . v ( t ) = 1 t V − 20 V − 10 V = t V − 30 V = ( t − 30 ) V Case (vi): 36 s < t ≤ 46 s The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 36 s , 6 V ) and ( 46 s , 6 V ) . Substitute 36 s for x 1 , 6 V for y 1 , 46 s for x 2 and 6 V for y 2 in equation (2). ( v ( t ) − 6 V ) = 6 V − 6 V 46 s − 36 s ( t − 36 s ) = 0 V 46 s − 36 s ( t − 36 s ) = 0 Simplify the above equation to find v ( t ) . v ( t ) = 0 V + 6 V = 6 V Case (vii): 46 s ≤ t < 50 s The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 46 s , 6 V ) and ( 50 s , 0 V ) . Substitute 46 s for x 1 , 6 V for y 1 , 50 s for x 2 and 0 V for y 2 in equation (2). ( v ( t ) − 6 V ) = 0 V − 6 V 50 s − 46 s ( t − 46 s ) = − 6 V 4 s ( t − 46 s ) = − 1.5 V ( t − 46 s ) = − 1.5 t V + 69 V Simplify the above equation to find v ( t ) . v ( t ) = − 1.5 t V + 69 V + 6 V = − 1.5 t V + 75 V = ( − 1.5 t + 75 ) V Case (viii): t > 50 s From the Figure 1, it is clear that the value of the voltage greater than the time period 50 s is 0 V . Therefore, the voltage function of the signal in Figure 1 is, v ( t ) = { 2.5 t V , 0 s ≤ t < 4 s 10 V , 4 s < t ≤ 8 s ( − 2.5 t + 30 ) V, 8 s ≤ t < 16 s − 10 V , 16 s < t ≤ 20 s ( t − 30 ) V , 20 s ≤ t < 36 s 6 V , 36 s < t ≤ 46 s ( − 1 b) To determine Find the energy delivered to the circuit element at t = 4 , 12 , 36 and 50 s

Electric Circuits, Student Value Edition Format: Unbound (saleable)

11th Edition

ISBN: 9780134747170

Author: NILSSON, James W.^riedel, Susan

Publisher: Prentice Hall

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Chapter 1, Problem 28P

To determine

Sketch the power versus time t plot for 0≤t≤50 s.

To determine

Find the energy delivered to the circuit element at t=4, 12, 36 and 50 s

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Chapter 1, Problem 27P

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Electric Circuits, Student Value Edition Format: Unbound (saleable)

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