Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

When two unknown resistors are connected in series with abattery, the battery de

ID: 1663142 • Letter: W

Question

When two unknown resistors are connected in series with abattery, the battery delivers 240 W andcarries a total current of 5.00 A. For the same total current,50.0 W is delivered when the resistors areconnected in parallel. Determine the values of the tworesistors. (the following is a different question all together) Two identical parallel-wired strings of 40 bulbs are connected to each other in series. If theequivalent resistance of the combination is 120.0 when it is connected across a potentialdifference of 120.0 V, what is the resistance of each individualbulb? When two unknown resistors are connected in series with abattery, the battery delivers 240 W andcarries a total current of 5.00 A. For the same total current,50.0 W is delivered when the resistors areconnected in parallel. Determine the values of the tworesistors. (the following is a different question all together) Two identical parallel-wired strings of 40 bulbs are connected to each other in series. If theequivalent resistance of the combination is 120.0 when it is connected across a potentialdifference of 120.0 V, what is the resistance of each individualbulb?

Explanation / Answer

        Firstproblem :          Power, P = I2 R          Case I:           P = 240W      I = 5 A           Rs = P /I2               = 9.6           R1 + R2 =9.6 .........(1)           Case II:           P = 50W      I = 5 A           Rp = P / I2               = 2          [ R1 * R2 / ( R1 +R2 ) ] = 2    ..............(2)           Solve (1)and (2) for R1 and R2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~        Secondproblem:          No. of bulbs ineach string, n = 40          Req = 120          Resistance of eachstring = 2 Req    ( since they are in parallel )                                               = 240         Let R = resistance ofeach bulb         n R = 240         Resistance of each bulb,R = 240 / 40                                                 = 6

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote