A 74.0 W and a 44.0 W resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.285 A. When the 44.0 W resistor is disconnected, the current from the battery drops to 0.150 A. Determine the emf of the battery.
2007-06-17 09:39:33 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Resistance of parallel combination = Rt
Rt = (74*44)/(74+44) = 27.6 Ω
Voltage across these resistor = 27.6*0.285 = 7.866 V
EMF = Ir + IRt
where r = internal resistance of battery
Considering the parallel combination;
E = 0.285r + 0.285*27.6
or, E = 0.285r + 7.866 -------(1)
For the 74Ω resistor only,
E = 0.150r + 0.150*44
or, E = 0.150r + 6.6 --------- (2)
Subtracting eq (2) from eq (1) gives...
[E-E] = [0.285r - 0.150r] - [7.866 - 6.6]
0 = 0.135r - 1.266
0.135r = 1.266
r = 9.4 Ω
Putting this into eq(1) gives:
E = (9.4*0.285) + (7.866) = 2.679 + 7.866
E = 10.545 V
2007-06-17 10:02:24 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I assume the resistors are 74.0 Ohms and 44.0 Ohms. Check this important detail first, before you are trying to solve the problem.
2007-06-17 16:51:13 · answer #2 · answered by Ernst S 5 · 0⤊ 1⤋
Use V = IR
1/Rt = 1/74 + 1/44
I = 0.285 A
2007-06-17 16:50:21 · answer #3 · answered by physandchemteach 7 · 0⤊ 1⤋
E.M.F=I(R+r)
I = Current
R = Resistance across curcuit
r = Internal Resistance
2007-06-17 16:54:39 · answer #4 · answered by | König | 2 · 0⤊ 0⤋
I do not do others home work. use your own brains. Teachers are there to help. don't be scared to ask them
2007-06-17 16:44:08 · answer #5 · answered by duster 6 · 0⤊ 2⤋