Growth of current in two different L−R circuits are depicted by the i−t graphs shown. Angle subtended by the curves with tine axis at time t=0 are also shown in the graph. τ1 and τ2 are time constant for the circuits 1 and 2 respectively. Then the value of τ1τ2 is
Circuit1:initially:Eo=L1di1dt=L1[tan30o]=L1√3=L1=√3Eo........(1)finally:io=EoR1⇒R1=Eoio.......(2)Circuit2:initiallyEo=L2di2dt⇒Eo=L2(tan60o)⇒L2=Eo√3..........(3)⇒R2=Eo2io..........(4)∴r1=L1R1=√3EoIoEo=√3ior2=L2R2=Eo2io√3.Eo=2io√3∴r1r2=√3×√32=32Hence,optionBiscorrectanswer.