A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 240 V to a primary coil of 280 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages? (b) If the maximum input current is 5.00 A, what are the maximum output currents (each used alone)?

A) i) at V_s = 5.6V

N_s = 6.53 turns

ii) at V_s = 12V, N_s = 14 turns

iii) at V_s = 480V, N_s = 560 turns

B) i) at V_s = 5.6V

I_s = 214.29A

ii) at V_s = 12V, I_s = 100A

iii) at V_s = 480V, N_s = 2.5A

Explanation:

A) The formula for calculating number of turns in the secondary coil is gotten from;

V_p/V_s = N_p/N_s

Making N_s the subject we have;

N_s =( V_s*N_p)/V_p

Where;

V_p is input voltage

V_s is output voltage

N_p is number of turns in primary coil

N_s is number of turns in secondary coil

We are given V_p = 240V and N_p = 280 turns

Thus;

i) at V_s = 5.6V,

N_s = (5.6*280)/240

N_s = 6.53 turns

ii) at V_s = 12V,

N_s = (12*280)/240

N_s = 14 turns

iii) at V_s = 480V,

N_s = (480*280)/240

N_s = 560 turns

B) The formula for calculating maximum output current in the secondary coil is gotten from;

I_s = (V_p*I_p)/V_s

Where;

I_s is maximum output current

V_p is input voltage

I_p is maximum input current

V_s is output voltage

We are given I_p = 5A

Thus;

i) at V_s = 5.6V,

I_s = (240*5)/5.6

I_s = 214.29 A

ii) at V_s = 12V

I_s = (240*5)/12

I_s = 100 A

iii) at V_s = 480V

I_s = (240*5)/480

I_s = 2.5 A

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

a. are