Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Chapter 16, Problem 1P

Find the total impedance of the parallel networks of Fig. 1663 in rectangular and polar form.

Chapter 16, Problem 1P, Find the total impedance of the parallel networks of Fig. 1663 in rectangular and polar form. Fig.

Fig. 1663

Expert Solution
Check Mark
To determine

(a)

Total impedance of the given network.

Answer to Problem 1P

Rectangular form: ZT=(941.17+j235.29)Ω

Polar form: ZT=970.14Ω14.03°

Explanation of Solution

Given:

The given network is:

  Introductory Circuit Analysis (13th Edition), Chapter 16, Problem 1P , additional homework tip  1

Calculation:

We can see from the given circuit that the resistor R and the inductive reactance XL are in parallel. Therefore, the total resistance of the network will be:

  ZT=R(jXL)ZT=( 1000)( j4000)1000+j4000ZT=j40000001000+j4000ZT=(941.17+j235.29)ΩZT=970.14Ω14.03°

Expert Solution
Check Mark
To determine

(b)

Total impedance of the given network.

Answer to Problem 1P

Rectangular form: ZT=(5679.23j1153.85)Ω

Polar form: ZT=5883.48Ω11.31°

Explanation of Solution

Given:

The given network is:

  Introductory Circuit Analysis (13th Edition), Chapter 16, Problem 1P , additional homework tip  2

Calculation:

We can see from the given circuit that the resistor R, capacitive reactance XC and the inductive reactance XL are in parallel. Therefore, the total resistance of the network will be:

  ZT=R(jXL)(jXC)ZT=11 j X L +1 j X C +1RZT=11 2000090°+1 1200090°+1 6000ZT=150× 10 690°+83.33× 10 690°+166.67× 10 6ZT=(5679.23j1153.85)ΩZT=5883.48Ω11.31°

Expert Solution
Check Mark
To determine

(c)

Total impedance of the given network.

Answer to Problem 1P

Rectangular form: ZT=j1224.32Ω

Polar form: ZT=1224.32Ω90°

Explanation of Solution

Given:

The given network is:

  Introductory Circuit Analysis (13th Edition), Chapter 16, Problem 1P , additional homework tip  3

Calculation:

We can see from the given circuit that two inductors and one capacitor are connected in parallel.

Therefore, we need to calculate their equivalent reactance first:

Inductive reactance of 40mH inductor:

  XL1=2πfL1XL1=(2π)(10× 103)(40× 10 3)XL1=2513.27Ω

Inductive reactance of 20mH inductor:

  XL1=2πfL2XL1=(2π)(10× 103)(20× 10 3)XL1=1256.64Ω

Capacitive reactance of 6nF inductor:

  XC=12πfCXC=1( 2π)( 10× 10 3 )( 6× 10 9 )XC=2652.58Ω

Therefore, the total resistance of the network will be:

  ZT=(jX L 1 )(jX L 2 )(jXC)ZT=11 j X L 1 +1 j X L 2 +1 j X C ZT=11 2513.2790°+1 1256.6390°+1 2652.5890°ZT=1398× 10 690°+377× 10 690°+795.78× 10 690°ZT=1816.78× 10 690°ZT=j1224.32ΩZT=1224.32Ω90°

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Chapter 16 Solutions

Introductory Circuit Analysis (13th Edition)

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