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?

Respuesta :

Answer:

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

Explanation:

WE HAVE FOUR CASES TO OBTAINED MAXIMUM POWER

we know that

Maximum current [tex]= \sqrt({P}{R})[/tex]

[tex]I =  \sqrt\frac{48}{2.4}v[/tex]

I= 4.47 volt

case 1 - arrange in series

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

[tex]power = 4.47^2 *7.2 = 143.86 w[/tex]

case 2 connected in parallel

[tex]R_EQUI = \frac{1}{\frac{1}{2.4}+\frac{1}{2.4}+\frac{1}{2.4}}[/tex]

[tex]R_EQUI = 0.80 \ohm[/tex]

power  = 0.80* 4.47^2 = 15.98 w

case 3 two in series and one in parallel

[tex]R _equi = \frac{(2.4+2.4)*2.4}{4.8+2.4} = 1.6\ohm[/tex]

[tex]power  = 4.47^2 *1.6 = 31.96 w[/tex]

case 4

one in series and two in parallel

[tex]R_equi = \frac{2.4*2.4}{2.4+2.4} +2.4 = 2.88 \ohm[/tex]

[tex]power = 4.47^2*2.88 = 57.54 w[/tex]