Question

A uniform wound solenoidal coil of self inductance `1.8xx10^(-4)` henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-volt battery of negligible resistance. The time constant for the current in the circuit is.......seconds and the steady state current through the battery is ..............amperes.

Answer 1

Correct Answer - A::C::D
The coil is broken into two idential coils.
`L_(eq)=(L//2xxL//2)/(L//2+L//2)=L/4=0.45xx10^(-4)H`,
`R_(eq)=(R//2xxR//2)/(R//2+R//2)=R/4=1.5(Omega)`
Time constant `(L_(eq))/(R_(eq))=(0.45xx10^(-4))/(1.5)=3.0xx10^(-4) s`
Steady current `I=(E)/(r_(eq))=15.1.5=10A`

Answer 2

Correct Answer - A::C::D
The coil is broken into two idential coils.
`L_(eq)=(L//2xxL//2)/(L//2+L//2)=L/4=0.45xx10^(-4)H`,
`R_(eq)=(R//2xxR//2)/(R//2+R//2)=R/4=1.5(Omega)`
Time constant `(L_(eq))/(R_(eq))=(0.45xx10^(-4))/(1.5)=3.0xx10^(-4) s`
Steady current `I=(E)/(r_(eq))=15.1.5=10A`