I1=R1F=212=6 AE=Ldtdl2+R2×I2I2=I0(1−e−t/tc)⇒Io=R2E=212=6 Atc=RL=2400×10−3=0.2I2=6(1−e−t/0.2) Potential drop across L=E−R2l2=12−2×6(1−e−bt)=12e−5t 
Directions: Question numbers 28,29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle ABCDA, as shown in the P−T diagram.