Question

# A 10.7-V battery, a 4.91-Ω resistor, and a 10.7-H inductor are connected in series. After the...

A 10.7-V battery, a 4.91-Ω resistor, and a 10.7-H inductor are connected in series. After the current in the circuit has reached its maximum value, calculate the following. (a) the power being supplied by the battery (b) the power being delivered to the resistor (c) the power being delivered to the inductor (d) the energy stored in the magnetic field of the inductor

(a) when current in the circuit has reached its maximum value

so

P=E2/R

P= 10.72 / 4.91

P= 23.31 W;

(b) when the current in the circuit has reached its maximum value

P=E2/R

P= 10.72 / 4.91

P= 23.31 W;

(c) After the current in the circuit has reached its maximum value :

so

as there is no potential difference between inductors ends

so P= 0

(d) After the current in the circuit has reached its maximum value this value is

I=E/R

I = 10.7/4.91 = 2.17 amp;

energy will be U=1/2 x L x I2

U= 1/2 x 10.7x 2.172 =25.19 J