Nothing

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

Vs = 12V; R1 = 300 Ω; R2 = 1.5 kΩ

IR1 = Vs /R1 = 12V/300 Ω = 0.04 A = 40 mA

IR2 = IT – IR1 = 50 mA – 40 mA = 10 mA

R2 = Vs/IR2 = 12V/10mA = 1.2 kΩ

So, Potentiometer setting required = 12. kΩ

18. For the circuit shown in Figure 5.33b, determine the change in VL that occurs when RL is changed from 470 Ω to 220 Ω.

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

Equivalent resistance of the three resistors in parallel = 1 / (1/120 + 1/100 + 1/150) = 40 Ω

Total resistance of circuit = 40 + 200 = 240 Ω

Voltage across RL = Voltage across parallel branch = 9 * (40/240) = 1.5 V

PL = V²/RL = 1.5^2 / 150 = 0.015 W (15 mW)

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

Equivalent resistance of parallel section = 1/(1/3 + 1/18) = 2.5714 kΩ

Total resistance of circuit = 12 + 2.5714 = 14.5714 kΩ

VL = 15*(2.5714/14.5714) = 2.647 V

VNL = 15*(3/(3+12)) = 3 V

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

Let the node between A and B be denoted by C.

Effective resistance between C and B = 1/ (1/240 + 1/360) = 144 Ω

Effective resistance between A and B = 1/ (1/330 + 1/(144+150)) = 155.48 Ω

Effective resistance between A and ground = 1/(1/120 + 1/(155.48+150)) = 86.156 Ω

So, VA = 6* 86.156/(100+86.156) = 2.777 V

So, VAB = 2.777* 155.48/(150+155.48) = 1.4134 V

So, V6 = 1.4134* 144/(150+144) = 0.692 V