My answers to these questions were incorrect, wondering what the correct solutio

Question

My answers to these questions were incorrect, wondering what the correct solution is.

A 15.0 Ohm resistor and a coil are placed in series with a 6.30 V battery which has negligible internal resistance. It has been connected to a switch which has been closed for a long time. At 2.00 ms after the switch was opened the current has decayed to 0.210 A. Calculate the inductance of the coil. Calculate the time constant of the circuit. How long after the switch is open will the current reach 1.00% of its original value?

Explanation / Answer

a) Io =E/R =6.3/15 =0.42 A => time constant T=L/R => I=Ioe^(-t/T)=>0.21 =0.42e^(-2/T) 0.5 =e^(-2/T)=> ln(0.5)=-2/T=> T=2.885 ms=> L/R =2.885*10^-3=> L=15*2.885*10^-3=> L=0.0433 H or 43.3mH=> b) T=L/R =0.0433/15 =2.89ms=> c) I=Ioe^(-t/T)=> 0.01Io =Ioe^(-t/2.89)=> ln(0.01) =-t/2.89=> t=13.31 ms

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