Question

In the circuit shown in figure the emf E of the battery is increased linearly from zero to 28 V in the
interval 0 <t<14s.
E
10 V
2.503
50
(A) The energy gained by the 10 V cell in the interval 0 <t<14s is 245 J
(B) The energy gained by the 10 V cell in the interval 0 <t<14s is OJ
(C) The time at which the 10 V cell begin to charge is t=75
(D) The time at which the 10 V cell begin to charge is t=14s​

Answer 1

Answer:

for loop AFDBEA:  EF−EE=i1(rF+rE+R)−Ri2

or 1−2=i1(1+2+2)−2i2⇒5i1−2i2=−1...(1)

for loop CGDBHC: EH−EG=i2(rH+rG+R)−Ri1

or 1−3=i2(1+3+2)−2i1⇒−2i1+6i2=−2....(2)

(1)×3⇒15i1−6i2=−3...(3)

(2)+(3),13i1=−5⇒i1