Three 100 Ohm resistors are connected as shown in the figure. The maximum power

Question

Three 100 Ohm resistors are connected as shown in the figure. The maximum power that can safely be delivered to any one resistor is 22.0 W. (a) What is the maximum potential difference that can be applied to the terminals a and b? V (b) For the voltage determined in part (a), what is the power delivered to each resistor? resistor on the left W resistor at the top of the loop W resistor at the bottom of the loop W (c) What is the total power delivered to the combination of resistors? W What is the equivalent resistance of the combination of identical resistors between points a and b in the figure below? R

Explanation / Answer

2. As per the given condition each resistor can safely bear a power input = 22.0 W.

Hence the maxiumum current allowed through any of the given resistor

( I ) = [ power / Resistance ]

=> I = [ 22.0 / 100 ] = 0.22 A

Hence considering the assumed circuit the resistance in series can have maximum current = 0.22 A, while each of the two connected in parallel will have currenet = (1/2) ( 0.22 ) = 0.11 A

Again total curcuit resistance = 100 + (100/2) = 150

Hence the maximum permissible voltage = 0.22 x 150 = 33 V........... Answer

Power delivered to the series resistor = (current)^2 x Resistance

=> (0.22)^2 x 100 = 4.84 W.... Answer

Power delivered to each of the resistances in parallel = (0.11)^2 x 100 = 1.21 W ........... Answer

Total Power delivered = 4.84 + (1/2) (4.84) = 7.26 Watt .... Answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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