What is the maximum potential difference that can be applied to the terminals a and b?

+3 votes

Three 100 Ω resistors are connected as shown in the figure. The maximum power that can safely be delivered to any one resistor is 22.5 W.

image
(a) What is the maximum potential difference that can be applied to the terminals a and b?
V

(b) For the voltage determined in part (a), what is the power delivered to each resistor?
resistor on the left W
resistor at the top of the loop W
resistor at the bottom of the loop W

(c) What is the total power delivered to the combination of resistors?
W

 

asked Feb 4, 2012 in Physics by yoshi ~Expert~ (919 points)
    

1 Answer

+3 votes
 
Best answer

The formula for power is:

P =  I^2 * R

The max power that can go to any resistor is 22.5 W.

so 

 22.5 W. = I^2 * R

and   I = squareroot (P/R) = squareroot (22.5 W / 100 Ω) =  0.4743 A (which is max current)

Now find the total resistance between a and b.

R total = 100  1/ (1/100 + 1/100)  = 150 Ω    (*notice how the resistors in parallel combine by taking the inverse of the sum of the recriprocal of the two resistors)

Max voltage = R total * max current

V = R*I  = (150 Ω)*(0.4743 A ) =  71.145 V

 

For Part B:

First caculate the total power of the circuit using the formula:

P = I*V = (0.4743 A)*(71.145 V) = 33.744 W

Now, to caculate the power delivered to each resistor you need to determine the percentage of total resistance each resistor makes up:

For example the first resistor is 100Ω out of the total of 150Ω, and so you will get:

(100/150)  Now multiply this by the total power and get:

 

(100/150)*33.744 = 22.496 W  for the resistor on the left.

 

Similarly the power in the other two resistors can be found in the same way, however, since they are connected in parallel, they will each receive the same power, but this power will be less then that of a resistor of same resistance connected in series.

One way to get this power, is to find the resistance of both resistors by adding them as follow:

1/ (1/100 + 1/100)  = 50Ω (*they are added this way because they are in parallel)

Now divide by 2, and you get 25Ω for each resistor.

 

Now set up the fraction, of 25/100 and multiply by the total power to get:

25/100  * 33.744 W = 5.6169W  (for each of the resistors in parallel)

 

answered Feb 8, 2012 by awesome ~Expert~ (1,479 points)
selected Feb 9, 2012 by yoshi

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