In the following figure (Figure 1) , R = 15.9ohms and the battery emf is 6.30V .

Question

In the following figure (Figure 1) , R = 15.9ohms and the battery emf is 6.30V . With switch S2 open, switch S1is closed. After several minutes, S1 is opened and S2is closed.

Part A

At 2.10ms after S1 is opened, the current has decayed to 0.278A . Calculate the inductance of the coil.

Express your answer to three significant figures and include the appropriate units.

Part B

How long after S1 is opened will the current reach 1.00% of its original value?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

At t=0-when s1 opened the initial current in the circuit is,

i(0) = V/R = 6.30/15.9=0.396

a)

   The current at any instant of time t is,

.i(t) = 0.396 * e^(-Rt/L)

0.278 = 0.396 * e^(-15.9(2.10 ms)/L)

=> L = 0.094 H

b)

The current at any instant of time t is,

i(t) = 0.396 * e^(-Rt/L)

=> 0.01 = e^(-15.9t/0.094)

=> t = 0.0273 sec

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.