Suppose a load with resistance R_L = 13.2 Ohm requires a current I_L = 115 mA to

Question

Suppose a load with resistance R_L = 13.2 Ohm requires a current I_L = 115 mA to flow through it. The only available current source provides I = 173 mA. The load may be connected to the current source in parallel or in series with another resistor, R. The design requirements can be met by placing R in series or in parallel with the load. Should it be connected in series or in parallel? What is the magnitude of R? What is the voltage drop across R_L in this configuration? If the power dissipated over R is considered wasted, what is the efficiency of this design?

Explanation / Answer

(A) here the resistance should be connected in series

Voltage should be same V = IR ===> 13.2 * 115 / 1000 = 1.518 V

Now When IL = 173 mA then ====> 1.518 = R * 173 / 1000

R = 8.774 ohm resistance should connected in series

(B ) Voltage Drop across = 1.518 Volt as calculated above

(c) Power Dissipated = I2R = 0.2625 w

Actual Power = IV =

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.