Asked by Brooke
Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 7.1Ω and 20.9 W, 29.5Ω and 11.4 W, and 13.6Ω and 12.5 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?
Answers
Answered by
Scott
I^2 = W / Ω = 11.4 / 29.5
calculate I
the resistors add in series
power = I^2 * total R
calculate I
the resistors add in series
power = I^2 * total R
Answered by
bobpursley
No, current will be limited by the lowest safe current a resistor can handle.
So do the above, but for EACH resistor. Now, pick the lowest current of them all, that will be safe current for all resistors.
Power = I^2 *total R
resistor1: I= for 7.1Ω and 20.9 W,, I= 4.12amp
resistor2:
I= 29.5Ω and 11.4 W=.62amp
resistor3:13.6Ω and 12.5 W
I=13.6Ω and 12.5 W = .962 amps
so the safe current is .62amps
Power=.62^1*total R
So do the above, but for EACH resistor. Now, pick the lowest current of them all, that will be safe current for all resistors.
Power = I^2 *total R
resistor1: I= for 7.1Ω and 20.9 W,, I= 4.12amp
resistor2:
I= 29.5Ω and 11.4 W=.62amp
resistor3:13.6Ω and 12.5 W
I=13.6Ω and 12.5 W = .962 amps
so the safe current is .62amps
Power=.62^1*total R
Answered by
Scott
missed part (a)
(a) voltage = [√(11.4/29.5)] * total R
(a) voltage = [√(11.4/29.5)] * total R
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