Question

Each of the three resistors in has a resistance of 2.4 Ω arranged in paralel and serial, and can dissipate a maximum of 48 W without becoming excessively heated. What is the maximum power the circuit can dissipate?

Answer 1

The maximum power the circuit can dissipate is 2880 W.

Step-by-step explanation:

To determine the maximum power the circuit can dissipate, we need to calculate the total resistance of the circuit. In parallel, the total resistance is the reciprocal of the sum of the reciprocals of the individual resistances. In this case, since all three resistors are the same, the total resistance would be 0.8 Ω.

Using the formula P = V²/R, where P is power, V is voltage, and R is resistance, we can calculate the maximum power dissipated by the circuit. With a maximum voltage of 48 V and a total resistance of 0.8 Ω, the maximum power the circuit can dissipate is:

P = (48 V)² / 0.8 Ω = 2880 W

Answer 2

maximum powr attained when resistor arrange in series and equal tp 143.86 w

Step-by-step explanation:

WE HAVE FOUR CASES TO OBTAINED MAXIMUM POWER

we know that

Maximum current
`= \sqrt({P}{R})`

`I =  \sqrt(48)/(2.4)v`

I= 4.47 volt

case 1 - arrange in series

equivalent resistance [/tex]= 3R = 3*2.4 = 7.2 \ohm[/tex]

`power = 4.47^2 *7.2 = 143.86 w`

case 2 connected in parallel

`R_EQUI = (1)/((1)/(2.4)+(1)/(2.4)+(1)/(2.4))`

`R_EQUI = 0.80 \ohm`

power = 0.80* 4.47^2 = 15.98 w

case 3 two in series and one in parallel

`R _equi = ((2.4+2.4)*2.4)/(4.8+2.4) = 1.6\ohm`

`power  = 4.47^2 *1.6 = 31.96 w`

case 4

one in series and two in parallel

`R_equi = (2.4*2.4)/(2.4+2.4) +2.4 = 2.88 \ohm`

`power = 4.47^2*2.88 = 57.54 w`