Ect Week 6 Assignment

Week 6 Assignment

Show all work

Chapter 5

15. Determine the potentiometer setting required to set the total circuit current in Figure 5.32a to 50 mA.

The potentiometer setting required is 1.2 kΩ.

I_1=V_S/R_1 =(12 V)/(300 Ω)=40 mA

I_2=I_T-I_1=50 mA-40 mA=10 mA

R_2=V_S/I_2 =(12 V)/(10 mA)=1.2 kΩ

Figure 5.32

(University, 11/2012, pp. 143)

38. Figure 5.43 shows a potentiometer wired as a current divider. Determine the values of RAB and RBC that will provide values of I1 = 4 mA and I2 = 6 mA.

RAB= 40 Ω, and RBC=60 Ω

Figure 5.43

(University, 11/2012, p. 146)

R_AB=I_2/I_T =(6 mA)/(10 mA)=40 Ω

R_BC=R_T-R_AB= 100 Ω-40Ω=60 Ω

Chapter 6

18. Determine the value of load power (PL) for the circuit shown in Figure 6.48b.

PL= 15 mW

FIGURE 6.48

R_EQ=1/(1/R_1 +1/R_L )=1/(1/(120 Ω+100 Ω+150 Ω))=1/(0.000833+.01+0.000667)=

1/.025=40 Ω

R_T=R_2+ R_EQ=200 Ω+40 Ω=240 Ω

V_EQ=V_S x R_EQ/R_T =9 V x (40 Ω)/(240 Ω)=9 Vx.1667=1.5 V

P_L=〖V_EQ〗^2/R_L =〖1.5 V〗^2/(150 Ω)=(2.25 V)/(150 Ω)=15 mW

20. Determine the values of VL and VNL for the circuit shown in Figure 6.49b.

VL = 2.25 V, and VNL = 3 V

FIGURE 6.49

R_EQ=1/(1/(R_2+R_L )) = 1/(1/(3 kΩ+18 kΩ))=1/(0.000333+0.0000556)=1/0.0003889

=2.5714 kΩ

R_T=R_1+ R_EQ=12 kΩ+2.573 kΩ=14.5714 kΩ

V_L= V_S x R_EQ/(R_EQ+R_T )=15 V x (2.5714 kΩ)/(2.5714 kΩ+14.5714 kΩ)=15 V x (2.5714 kΩ)/(17.1428 kΩ)=

15 V x .15=2.25 V

V_NL= V_S x R_2/(R_1+R_2 )=15 V x (3 kΩ)/(12 kΩ+3 kΩ)=15 V x (3 kΩ)/(15 kΩ)=15 V x.2=3V

38. Determine the value of V6 for the circuit shown in Figure 6.56.

V6 = 1.9 V

FIGURE 6.56

R_EQ1=1/(1/R_5 +1/R_6 ) =1/(1/(240 Ω+360 Ω))=1/(0.004167+0.002778)=1/0.006945=143.99 Ω

R_EQ2=1/(1/(R_3+(R_4+R_EQ1))) =1/(1/(330 Ω+(150 Ω+143.99 Ω)))=1/(0.00303+0.0034)=

1/.00643=155.52 Ω

R_EQ3=1/(1/(R_2+(R_EQ2+R_L))) =1/(1/(120 Ω+(155.52 Ω+150 Ω) ))=1/(0.00833+0.003273)=

1/0.0116=86.21 Ω

R_T=R_1+R_EQ3=100 Ω+86.21 Ω=186.21 Ω

V_6= V_S x R_EQ3/(R_EQ3+R_T )=6 V x (86.21 Ω)/(86.21...