ENGR 2219 - Linear Circuits I

Homework Answers

 

Homework 1

2.4)  P_dev = 300 W

2.8)  The interconnection is invalid

2.18)  i_b = 0.5 A, v₀ = 40 V, p₄Ω = 25 W, p_20Ω = 80 W, p_80Ω = 20 W, P_dev = 125 W

2.19)  i₁ = 2.4 A, i₂ = 1.6 A, v₀ = 192 V, P_dev = P_abs = 768 W

 

Homework 2

3.7)  (a) 16Ω , (b) 6 kΩ

3.10)  480 W

3.28) i₀ = 1.44 A, P₁₅Ω = 345.6 W

 

Homework 3

4.12) v₁ = 100 V, v₂ = 20 V 

4.20) P_dev = P_abs = 165 W

4.32) P_dev = P_abs = 1140 W  

4.43) P_dev = P_abs = 5157 W

4.59) (a) 3 mA

4.63) v_Th = 48 V, R_Th = 16 Ω

 

Homework 4

(a) I = 6 mA, V = 0

(b) I = 0, V = 0

(c) I = 3.33 mA, V = 3.33 V

(d) I = 0, V = 6.67 V

(e) I = 16.67 mA, V = 20 V

 

Homework 5

(1) I_C = 1.93 mA, V_CE = 10.35 V

(2) I_C = 9.9 mA, V_CE = -0.2 V

(3) I_C = 0 mA, V_CE = 10 V

(4) (a) I_C = 3.61 mA, V_CE = 1.84 V, (b) I_C = 3.1 mA, V_CE = 0.2 V

 

Problem 2

I_DQ = 2.82 mA, V_DSQ = 11.54 V

 

Homework 6

5.2)

(a) -12 V 

(b) -18 V

(c) 10 V

(d) -14 V

(e) 18 V

(f) 2.8125 to 7.3125 V   

      

5.3) -3.1 mA

 

5.12)   

(a) Summing Amplifier 

(b) -6 V

(c) -7.5 to 1.5 V

 

5.17)

(a) Non-inverting Amplifier 

(b) v₀ = 3.75v_s V

(c) -7.5 to 1.5 V

 

5.24) 20 kΩ

 

Homework 7

6.3) iL(t) = 5-2.5e^-3t A

 

6.4)

(a) 2e-^5t(1-5t) mV

(b) 735.76 µW

(c) absorbing

(d) 73.58 µJ

(e) 108.27 µJ

 

6.14)

0 < t < 1:  i_C(t) = 60t² µA

1 < t < 2:  i_C(t) = -60(2-t²) µA

 

6.15)

(a) -(200 x 10³)t + 40 V

(b) (4 x 10⁵)t - 20 V

(c) 100V

 

Homework 8

7.2)

(a) 0.2 mA

(b) i₁ = 0.2 mA, i₂ = -0.2 mA

(c) 0.2e^-1000000t

(d) -0.2e^-1000000t

 

7.24)

(a) 39.6e^-2000t mA

(b) 14.05 %

 

7.50)

(a) 40 - 40e^-5000t V

(b) 2 - e-5000t mA

(c) 2 + 4e^-5000t mA

(d) 8 - 4e^-5000t mA

(e) 6 mA

 

Homework 9

8.6)

(a) i_R(0) = 45 mA, i_L(0) = -30 mA, i_C(0) = -15 mA

(b) 70e^-10,000t + 20e^-40,000t V

(c) -28e^-10,000t - 2e^-40,000t mA

 

8.42)

(a) 300 V

(b) -12,000 V/s

(c) 300e^-80tcos60t + 200e^-80tsin60t V

 

Homework 10

9.23) 30 - j40 Ω

 

9.29)

(a) I_b = 20 + j20 mA = 28.28 Ð 45° mA, I_c = 100 + j100 mA = 141.42 Ð 45° mA, V_g = 7.3 + j4.1 mA = 8.37 Ð 29.32° Volts

(b) i_b = 28.28cos(800t + 45°) mA, i_c = 141.42cos(800t + 45°) mA, v_g = 8.37cos(800t + 29.32°) Volts

 

9.39)

(a) 0.8 H and 0.2 H

(b) i_g = 20cos(10000t) mA for 0.8 H, i_g = 40cos(10000t) mA for 0.2 H

 

9.41) I_N = 8 Ð -36.87° A, R_N = 50 - j25 Ω

 

 

 

 

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