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```        The electric current in a discharging R-C circuit is given by i = i0 e-t/RC where i0,  R and C are constant parameters and t is time. Let i0 =2.00 A, R = 6.00×105 Ω and c = 0.500 μF.  (a) Find the current at t = 0.3 s.
(b) Find the rate of change of current at t = 0.3 s.
c) Find approximately the current at t = 0.31 s.
```
4 years ago

Jitender Pal
365 Points
```							Sol. Equation i = i0e-t/RC
I0 = 2A, R = 6 x 10-5 Ω, C = 0.0500 x 10-6 F = 5 x 10-7 F
(a) i = 2 x e ((-0.3)/(6 x 0^3 x 5 x 〖10〗^(-7) )) = 2 x e((-0.3)/0.3) = 2/e amp.
(b) di/dt = (-i_0)/RC e-t/RC when t = 0.3 sec ⇒ di/dt = - 2/0.30 e(-0.3/0.3) = (-20)/3e Amp/Sec
(c) At t = 0.31 sec, I = 2e(-0.3/0.3) = 5.8/3e Amp

```
4 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions