ECT122 Wk6 Assignment
Homework Week 6
Chapter 5 Practice Problems
15. Determine the potentiometer setting required to set the total circuit current in Figure 5.32a to 50mA.
I1=Vs/R1= 12V/300Ω= 0.04 x 10^-3= 40mA
I2=It – I1= 50mA – 40mA= 0.01 x 10^-3= 10mA
R2=Vs/I2= 12V/10mA= 1200 x 10^3= 1.2kΩ
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= (4mA/10mA) x 100Ω= .4 x 100Ω= 40 Ω
Rbc= (6mA/10mA) x 100Ω= .6 x 100Ω= 60Ω
Chapter 6 Practice Problems
18. Determine the value of load power (PL) for the circuit shown in Figure 6.48b.
Ra= R1+R2= 120Ω+200Ω= 320Ω
Rt= Ra//R3//RL= 320Ω//100Ω//150Ω= 50.5Ω
It= Vs/Rt= 9V/50.5Ω= 0.17821 x 10^-3= 178.2mA
PL= VL^2/RL= 9V^2/150Ω= 0.54 x 10^-3= 540mW
20. Determine the values of VL and VNL for the circuit shown in Figure 6.49b.
Req= R2//RL= 3kΩ//18kΩ= 2571 x 10^3= 2.571kΩ= 2.57kΩ
VL= V1 x Req/R1+Req= 15V x 2.57kΩ/(12kΩ+2.57kΩ)= 15V x 2.57kΩ/ 14.57kΩ= 2.6V
VNL= Vs x R2/R1+R2= 15V x 3kΩ/(12kΩ+3kΩ)= 15V x 3kΩ/15kΩ= 3V
38. Determine the value of V6 for the circuit shown in Figure 6.56.
Req= R5//R6= 240Ω//360Ω= 144Ω
V6= Vs (Req/R4+Req)= 6V x 144Ω/150Ω+144Ω= 6V x 144Ω/294Ω= 2.94V
40. Calculate the total resistance in the circuit shown in Figure 6.58.
R14= R4//R1=10kΩ//22kΩ= 6875 x 10^3=6.875kΩ = 6.9kΩ
R39= R9//R3=6.8kΩ//4.7kΩ= 2779 x 10^3= 2.779kΩ= 2.8kΩ
R26= R6//R2= 3.3kΩ//11kΩ= 2538 x 10^3= 2.5kΩ
R839= R8//R39= 11kΩ//2.8kΩ= 2231 x 10^3= 2.23kΩ
Req1= R14 +R26 + R398= 6.9kΩ+2.5kΩ+2.23kΩ= 11630 x 10^3= 11.63kΩ
Req2= R7//Req1= 8.2kΩ//11.63kΩ= 4809 x 10^3= 4.8kΩ
Rt= R10+Req2 = 10kΩ+4.8kΩ= 14800 x 10^3= 14.8kΩ
41. The alarm trigger circuit in Figure 6.59 has the following characteristics: a. It activates the alarm when the voltage at input Y (VY) equals the voltage at input X(VX). b. VY is provided by a digital thermometer with an output of 0.05 V/°C. c. VX is a reference voltage that is determined by the setting of R2. d. The X and Y
Chapter 5 Practice Problems
15. Determine the potentiometer setting required to set the total circuit current in Figure 5.32a to 50mA.
I1=Vs/R1= 12V/300Ω= 0.04 x 10^-3= 40mA
I2=It – I1= 50mA – 40mA= 0.01 x 10^-3= 10mA
R2=Vs/I2= 12V/10mA= 1200 x 10^3= 1.2kΩ
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= (4mA/10mA) x 100Ω= .4 x 100Ω= 40 Ω
Rbc= (6mA/10mA) x 100Ω= .6 x 100Ω= 60Ω
Chapter 6 Practice Problems
18. Determine the value of load power (PL) for the circuit shown in Figure 6.48b.
Ra= R1+R2= 120Ω+200Ω= 320Ω
Rt= Ra//R3//RL= 320Ω//100Ω//150Ω= 50.5Ω
It= Vs/Rt= 9V/50.5Ω= 0.17821 x 10^-3= 178.2mA
PL= VL^2/RL= 9V^2/150Ω= 0.54 x 10^-3= 540mW
20. Determine the values of VL and VNL for the circuit shown in Figure 6.49b.
Req= R2//RL= 3kΩ//18kΩ= 2571 x 10^3= 2.571kΩ= 2.57kΩ
VL= V1 x Req/R1+Req= 15V x 2.57kΩ/(12kΩ+2.57kΩ)= 15V x 2.57kΩ/ 14.57kΩ= 2.6V
VNL= Vs x R2/R1+R2= 15V x 3kΩ/(12kΩ+3kΩ)= 15V x 3kΩ/15kΩ= 3V
38. Determine the value of V6 for the circuit shown in Figure 6.56.
Req= R5//R6= 240Ω//360Ω= 144Ω
V6= Vs (Req/R4+Req)= 6V x 144Ω/150Ω+144Ω= 6V x 144Ω/294Ω= 2.94V
40. Calculate the total resistance in the circuit shown in Figure 6.58.
R14= R4//R1=10kΩ//22kΩ= 6875 x 10^3=6.875kΩ = 6.9kΩ
R39= R9//R3=6.8kΩ//4.7kΩ= 2779 x 10^3= 2.779kΩ= 2.8kΩ
R26= R6//R2= 3.3kΩ//11kΩ= 2538 x 10^3= 2.5kΩ
R839= R8//R39= 11kΩ//2.8kΩ= 2231 x 10^3= 2.23kΩ
Req1= R14 +R26 + R398= 6.9kΩ+2.5kΩ+2.23kΩ= 11630 x 10^3= 11.63kΩ
Req2= R7//Req1= 8.2kΩ//11.63kΩ= 4809 x 10^3= 4.8kΩ
Rt= R10+Req2 = 10kΩ+4.8kΩ= 14800 x 10^3= 14.8kΩ
41. The alarm trigger circuit in Figure 6.59 has the following characteristics: a. It activates the alarm when the voltage at input Y (VY) equals the voltage at input X(VX). b. VY is provided by a digital thermometer with an output of 0.05 V/°C. c. VX is a reference voltage that is determined by the setting of R2. d. The X and Y