a) To determine Sketch the power versus time t plot for 0 ≤ t ≤ 80 ms . Explanation Given data: Refer to the given plots in the problem. Calculation: The given current 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 i ( t ) for y in equation (1). ( i ( 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 i ( t ) (3) Here, v is the value of the voltage, and i ( t ) is the value of the current. Refer to Figure 1, split up the time period as five divisions 0 ms ≤ t < 10 ms , 10 ms < t ≤ 30 ms , 30 ms ≤ t < 40 ms , 40 ms < t ≤ 60 ms and t > 60 ms to find the respective current value. Case (i): 0 ms ≤ t < 10 ms The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 0 ms , 0 mA ) and ( 10 ms , 25 0 mA ) . Substitute 0 ms for x 1 , 0 mA for y 1 , 10 ms for x 2 and 25 0 mA for y 2 in equation (2). ( i ( t ) − 0 mA ) = 25 0 mA − 0 mA 10 ms − 0 ms ( t − 0 ms ) = 25 0 mA 10 ms t = 25 0 × 10 − 3 A 10 × 10 − 3 t { ∵ 1 m = 10 − 3 } = 25 t A Simplify the above equation to find i ( t ) . i ( t ) = 25 t A + 0 mA = 25 t A Case (ii): 10 ms < t ≤ 30 ms The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 10 ms , 25 0 mA ) and ( 30 ms , − 25 0 mA ) . Substitute 10 ms for x 1 , 25 0 mA for y 1 , 30 ms for x 2 and − 25 0 mA for y 2 in equation (2). ( i ( t ) − 25 0 mA ) = − 25 0 mA − 25 0 mA 30 ms − 10 ms ( t − 10 ms ) = − 500 mA 20 ms ( t − 10 ms ) = − 500 × 10 − 3 A 20 × 10 − 3 ( t − ( 10 × 10 − 3 ) ) { ∵ 1 m = 10 − 3 } = − ( 25 ( t − 0.01 ) ) A Simplify the above equation to find i ( t ) . i ( t ) = − 25 t A + 0 .25 A + ( 25 0 × 10 − 3 ) A { ∵ 1 m = 10 − 3 } = − 25 t A + 0 .25 A + 0.25 A = − 25 t A + 0 .5 A = ( 0.5 − 25 t ) A Case (iii): 30 ms ≤ t < 40 ms The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 30 ms , − 25 0 mA ) and ( 40 ms , − 25 0 mA ) . Substitute 30 ms for x 1 , − 25 0 mA for y 1 , 40 ms for x 2 and − 25 0 mA for y 2 in equation (2). ( i ( t ) − ( − 25 0 mA ) ) = − 25 0 mA − ( − 25 0 mA ) 40 ms − 30 ms ( t − 30 ms ) ( i ( t ) + 25 0 mA ) = − 25 0 mA + 25 0 mA 10 ms ( t − 30 ms ) = 0 10 ms ( t − 30 ms ) = 0 Simplify the above equation to find i ( t ) . i ( t ) = 0 − 250 mA = − 250 mA Case (iv): 40 ms < t ≤ 60 ms The two points ( x 1 , y 1 ) and ( x 2 , y 2 ) are ( 40 ms , − 25 0 mA ) and ( 60 ms , 25 0 mA ) . Substitute 40 ms for x 1 , − 25 0 mA for y 1 , 60 ms for x 2 and 25 0 mA for y 2 in equation (2). ( i ( t ) − ( − 25 0 mA ) ) = 25 0 mA − ( − 25 0 mA ) 60 ms − 40 ms ( t − 40 ms ) ( i ( t ) + 25 0 mA ) = 25 0 mA + 25 0 mA 20 ms ( t − 40 ms ) = 500 × 10 − 3 A 20 × 10 − 3 ( t − ( 40 × 10 − 3 ) ) { ∵ 1 m = 10 − 3 } = 25 ( t − ( 40 × 10 − 3 ) ) A Simplify the above equation to find i ( t ) . i ( t ) = 25 t A − 1 A − ( 250 × 10 − 3 ) A { ∵ 1 m = 10 − 3 } = 25 t A − 1 A − 0.25 A = 25 t A − 1.25 A = ( 25 t − 1.25 ) A Case (v): t > 60 ms From the Figure 1, it is clear that the value of the current greater than the time period 60 ms is 250 mA . Therefore, the current function of the signal in Figure 1 is, i ( t ) = { 25 t A , 0 ms ≤ t < 10 ms ( 0 b) To determine Find the energy delivered to the circuit element at t = 10 , 30 and 8 0 ms

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Author: James W. Nilsson, Susan Riedel

Publisher: PEARSON

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Chapter 1, Problem 27P

To determine

Sketch the power versus time t plot for 0≤t≤80 ms.

To determine

Find the energy delivered to the circuit element at t=10, 30 and 80 ms

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Chapter 1, Problem 26P

Chapter 1, Problem 28P

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