(a) Explanation Given data: The circuit diagram is given in Figure 17.80. The value of resistance ( R ) is 2 k Ω . The value of capacitor is 100 μ F . The applied current is, i ( t ) = 20 + 16 cos ( 10 + 45 ° ) + 12 cos ( 20 t − 60 ° ) mA (1) Formula used: Write the Ohm’s law formula to find the voltage. V = I R (2) Here, V is the voltage, I is the current, and R is the resistance. Write the expression for capacitive impedance. Z c = 1 j ω n C (3) Here, C is the capacitor, and ω n is the frequency. Write the general expression for voltage in amplitude-phase form. v ( t ) = V dc + ∑ n = 1 ∞ V n cos ( n ω 0 t − θ n ) (4) Here, V dc is the DC component of voltage, V n is the AC component of voltage, n is the number of components, ω 0 is the frequency, and θ n is the phase angle between the voltages. Write the general expression for current in amplitude-phase form. i ( t ) = I dc + ∑ m = 1 ∞ I m cos ( m ω 0 t − ϕ m ) (5) Here, I dc is the DC component of current, I m is the AC component of current, m is the number of components, ω 0 is the frequency, and ϕ m is the phase angle between the currents. Calculation: Expand the equation (5) as follows. i ( t ) = I dc + ∑ m = 1 2 I m cos ( m ω 0 t − ϕ m ) i ( t ) = I dc + I 1 cos ( ω 0 t − ϕ 1 ) + I 2 cos ( 2 ω 0 t − ϕ 2 ) (6) Compare equations (1) and (6). I dc = 20 mA I 1 = 16 mA I 2 = 12 mA ω 0 = 10 In the given figure, while calculating the DC component values, the capacitor acts like an open circuit. Therefore, DC voltage depends upon the value of current and resistor. Modify equation (2) to find V dc . V dc = I dc R Substitute 2 k Ω for R and 20 mA for I dc to find V dc . V dc = ( 20 mA ) ( 2 k Ω ) = ( 20 × 10 − 3 A ) ( 2 × 10 3 Ω ) { ∵ 1 mA = 1 × 10 − 3 A ∵ 1 k Ω = 1 × 10 3 Ω } = 40 V Find the value of AC component of voltages. Write the expression for ω n . ω n = m ω 0 Substitute 10 for ω 0 . ω n = 10 m Substitute 10 m for ω n and 100 μ F for C in equation (3) to find capacitive impedance. Z c = 1 j ( 10 m ) ( 100 μ F ) = 1 j ( 10 m ) ( 100 × 10 − 6 F ) { ∵ 1 μ F = 1 × 10 − 6 F } = 1 j 0.001 m Z c = ( 1 j m ) k Ω In the given figure, capacitor and resistor are connected in parallel combination. Find the total impedance of the circuit. Z m = R ∥ Z c Substitute 2 k Ω for R and ( 1 j m ) Ω for Z c to find Z . Z m = ( 2 k Ω ) ∥ ( ( 1 j m ) k Ω ) = ( 2 k Ω ) ( ( 1 j m ) k Ω ) ( 2 k Ω ) + ( ( 1 j m ) k Ω ) = ( 2 j ) ( 2 + 1 j m ) Ω = ( − j 2 m ) ( 2 − j m ) Ω Simplify the expression as follows (b) To determine Find the value of average power dissipated in the resistor.

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ISBN: 8220102801448

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Chapter 17, Problem 48P

(a)

To determine

Find the value of v(t) for the circuit given in Fig 17.80.

(b)

To determine

Find the value of average power dissipated in the resistor.

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Chapter 17, Problem 47P

Chapter 17, Problem 49P

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The values of the elements in the circuit given in the figure are given below. Find the maximum average power that can be transferred to the ZL load. Vg=5cos(10000t) VoltR=38 kilo ohmC=35 nano faradL=150 milli henry

Chapter 17 Solutions

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS

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Find the value of v ( t ) for the circuit given in Fig 17.80.

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Find the value of v(t) for the circuit given in Fig 17.80.

Find the value of average power dissipated in the resistor.

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Chapter 17 Solutions