Question

A circuit for light bulbs in a car has a battery with a terminal voltage of 24.5 V. The internal resistance of this battery is 0.66 Ohms. All bulbs have the same efficiency. The first bulb of this circuit has 15 Ohms and is in series with the battery. After this there are two branches. One branch has the third bulb with 5 Ohms and the fourth bulb with 7 Ohms. The third and fourth bulbs are in series with each other. The other branch has the second and fifth through eighth bulbs. The resistances of all these bulbs in Ohms are R2=9, R5=27, R6=48, R7=68, and R8=36, respectively. The fifth and sixth bulbs are in parallel with each other but in series with all other bulbs in the second branch, which are all in series with each other. Assume all questions begin with this circuit.The permittivity of free space is 8.85418782E(-12) F/m and the permeability of free space is 1.25663706E(-6) T*m/A.

You replace all the bulbs in the original circuit with capacitors of the same capacitance (in Farads) as the magnitude of the resistance in the bulbs. You then charge this circuit with the car battery. The battery is removed and the capacitors are used to run the car stereo which has a power consumption of 330 Watts. How long, in seconds, will the stereo run assuming a constant power consumption?

Answer 1

battery voltage, V = 24.5 V ( termianl voltage)
internal resistance of this battery, r = 0.66 ohms
efficiency of all the bulbs be n
given R1 in series with battery, R1 = 15 ohms
R3 = 5 ohms
R4 = 7 ohms
R3 and R4 are in series

R2 = 9 ohm
R5 = 27 ohms
R6 = 48 ohms
R7 = 68 ohms
R8 = 36 ohms
R5 and R6 are in parallel, and these are in series with R2, R7, R8

so, when these are replaced with capacitors, of same capacitance
capacitors in parallel are added algebriacally, in series are added like resistors in p[arallel]

so, C5 and C6 give, C' = C5 + C6 = 75 F
C2, C7, C8 give C" = (1/9 + 1/68 + 1/36) ^-1 = 6.510 F
C' and C" give Cl = C'C"/(C' + C") = 5.9906 F

C3 and C4 give, Co = C3*C4/(C3 + C4) = 2.9166 F

Co and Cl give Cm = Co + Cl = 8.90722

anf Ceff = Cm*C1/(Cm + C1) = 5.588 F
energy stored in this capacitor E = 0.5*Ceff*V^2 = 1677.298 J
speaker power = 330 W
so time for which the spe3akers work = t
330*t = 1677.298
t = 5.08 seconds