Ect 122 Wk5

Chapter Practice Problems

3. Calculate the component voltages and branch currents for the circuit shown in Figure 6.40, along with the values of IT and RT.

Ra = 10kΩ + 3kΩ = 13kΩ Rb = 4.7kΩ + 3.3kΩ = 8kΩ Vab = (26V) 4.9kΩ/2kΩ + 4.9kΩ = 18.46V – branch volts

Ia = 26V/13kΩ = 2mA Ib = 26V/8kΩ = 3.25mA Rt = 2kΩ + 4.9kΩ = 6.9kΩ

V1 = 2mA * 10kΩ = 20V V2 = 2mA * 3kΩ = 6V It = 26V/6.9kΩ = 3.8mA

V4 = 3.25mA * 4.7kΩ = 15.3V V5 = 3.25mA * 3.3kΩ = 10.7V

Req = (13kΩ)(8kΩ)/ 13kΩ + 8kΩ = 104000000Ω/ 21kΩ = 4.9kΩ

5. Calculate the component currents and loop voltages for the circuit shown in Figure 6.42, along with the values of IT and RT.

Ra = (220Ω)(100Ω)/220Ω + 100Ω = 68.75Ω Rb = (510Ω)(470Ω)/510Ω + 470Ω = 244.6Ω

Va = VsRaRa+Rb = (6V) 68.75 Ω/(68.75 Ω + 244.6 Ω) = 1.32V

Vb = (6V)244.6 Ω/(68.75 Ω + 244.6 Ω) = 4.68V

I1 = 1.32V/220Ω = 6mA I2 = 1.32V/100 Ω = 13.2mA

I3 = 4.68V/510 Ω = 9.2mA I4 = 4.68V/470 Ω = 9.9mA

Rt = 68.75 Ω + 244.6 Ω = 313.35 Ω

It = 6V/313.35 Ω = 19.1mA

8. Calculate the component currents and loop voltages for the circuit shown in Figure 6.45, along with the values of IT and RT.

Ra = (5.6k Ω)(12k Ω)/5.6k Ω + 12k Ω = 3.8k Ω Rb = (2k Ω)(3.6k Ω)/2k Ω + 3.6k Ω = 1.3k Ω

Rc = (7.5k Ω)(10k Ω)/7.5k Ω + 10k Ω = 4.3k Ω

Va = (18V) 3.8k Ω/3.8k Ω + 1.3k Ω + 4.3k Ω = 7.3V Vb = (18V) 1.3k Ω/ 3.8k Ω + 1.3k Ω + 4.3k Ω = 2.5V

Vc = (18V) 4.3k Ω/3.8k Ω + 1.3k Ω + 4.3k Ω = 8.2V

I1 = 7.3V/5.6k Ω = 1.3mA I2 = 7.3V/12k Ω = 608.3µA I3 = 2.5V/2k Ω = 1.25mA

I4 = 2.5V/3.6k Ω = 694.4µA I5 = 8.2V/7.5k Ω = 1.1mA I6 = 8.2V/10k Ω = 820µA

Rt = 3.8k Ω + 1.3k Ω + 4.3k Ω = 9.4k Ω

It = 18V/9.4k Ω = 1.9mA

14. Determine the value of load voltage (VL) for the circuit shown in Figure 6.46b.

Req = (R2 + R3) || RL = (33 Ω + 120 Ω)|| 39 Ω = 31.1 Ω

Veq = (12V) 31.1 Ω/ 31.1 Ω + 22 Ω + 51 Ω = 3.6V

Voltage across parallel branches. VL = 3.6V

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

Req1= (15 Ω)(10V)/15 Ω + 10 Ω = 6 Ω...