475) help in next few hours please!!!! A battery, whose e. m. f. (constant) is a

Question

475) help in next few hours please!!!!

A battery, whose e. m. f. (constant) is and whose internal resistance is r (constant), is connected to an external circuit consisting of a variable resistance, R. Find the value of R that causes the output power from the battery to be maximum (i.e., the power developed in R), and show that this maximum power occurs when the current in the circuit is one-half the short-circuit current from the battery. Show that the maximum power developed in R is ^2/4R. The current in a wire varies according to the relation i = 20 sin (377t), where i is in Ampere t in seconds, and (377t) in radians. How many Coulombs pass a cross section of the wire in the time interval between t = 0 and t = 1/120 sec? In the interval between t = 0 and t = 1/60 sec? What is the average current in each of the intervals above? What is the rms current in each interval? (Work out the integral.) The diagram shows a graph of a half-wave rectified alternating current. During the intervals from t = 0 to t = T/2, and t = T to t = 3T/2, etc., the current is given by I = I_m sin(2 pi t/T). During other intervals, the current is zero. Find the average current during the interval 0 to T. Find the rms current during the same interval.

Explanation / Answer

475)   a)    The equivalent resistance of the circuit is   r + R

     => Current , I = E/(r + R)

=> The power output of the emf   = EI

                                                     =   E2/(r + R)

=>   The power dissipated as heat by the internal resistance of the battery   = E2r/(r + R)2

=>    power transferred to the load ,P =   E2R/(r + R)2

The function P increases monotonically from zero for increasing R , and attains a maximum value at   R = r

=>    value of R that causes power output from battery maximum = r

=>   short circuit current =   E/r

=>     current in circuit for maximum power = E/(r + r)

                                                                    = E/2r

=>   current in circuit for maximum power is one half of   short circuit current   .

b)     maximum power is dissipated when r = R

=>     maximum power dissipated in R = E2R/(R + R)2

                                                                         =    E2R/4R2

                                                                         =    E2/4R

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