ch30-p059 - t = 0, just after the switch is closed. Now,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
59 (a) Assume i is from left to right through the closed switch. Let i 1 be the current in the resistor and take it to be downward. Let i 2 be the current in the inductor and also take it to be downward. The junction rule gives i = i 1 + i 2 and the loop rule gives i 1 R L ( di 2 /dt ) = 0. Since di/dt =0 , the junction rule yields ( di 1 /dt )= ( di 2 /dt ). Substitute into the loop equation to obtain L di 1 dt + i 1 R =0 . This equation is similar to Eq. 30 44, and its solution is the function given as Eq. 30 45: i 1 = i 0 e Rt/L , where i 0
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: t = 0, just after the switch is closed. Now, just after the switch is closed, the inductor prevents the rapid build-up of current in its branch, so at that time, i 2 = 0 and i 1 = i . Thus i = i , so i 1 = ie Rt/L and i 2 = i i 1 = i 1 e Rt/L . (b) When i 2 = i 1 , e Rt/L = 1 e Rt/L , so e Rt/L = 1 2 . Take the natural logarithm of both sides and use ln(1 / 2) = ln 2 to obtain ( Rt/L ) = ln 2 or t = L R ln 2 ....
View Full Document

Ask a homework question - tutors are online