Question

Hi can someone show how is the integration and the time derivative done (This is a topic on RC circuits)

We can derive general expressions for charge q and current i as functions of time. With our choice of the positive direction for current (Fig. 26.20b), i equals the rate at which positive charge arrives at the left-hand (positive) plate of the capacitor, so i = dq/dt. Making this substitution in Eq. (26.10), we have do E dr R RC = P(q - CE) We can rearrange this to da 4 - CE dt RC and then integrate both sides. We change the integration variables to q' and ' so that we can use q and for the upper limits. The lower limits are q' = 0 and 1 = 0; 50*.40E -- SOS When we carry out the integration, we get I( E) - RC Exponentiating both sides (that is, taking the inverse logarithm) and solving for q. we find 4-CE /RC -CE R-C circuit charging capacitor: Capacitor charge Capacitance Final capacitor charge = CE q = cɛ(1 – e IRC) = 0 (1 – et/BC) Battery emf Time since Resistance switch closed (26.12) The instantaneous current i is just the time derivative of Eq. (26.12): Current Battery emf R-C circuit, charging capacitor: Time since switch closed Initial current RC = le RC = ER (26.13) Ru Rate of change of Resistance Capacitance capacitor charge

Answer 1

de Rc (9-cE) Divide both side by (9-CE) we get da dt (A-CE) = multiplying both sides by at we get, da (ACE) = - Ride P1100 Now integrating both sides with given limits. Saco do => [lin (9-CE)]! - F C :S3 da= sna) - [In (9-CE) - Jn Co-ce) = Ei el In (9-C E) - In (-CE) = - ť 7 in (966) - ... [idna- snb = 10 (39] Rc Now eseponentiating both sides 9-CE = HRC ... (: anze_re) - - CE e RCA CE 10- CEC-et leg) (:9-C1) REDMI NOTE 7 PRO

:: 9 = CE (1-ēt/RC) Taking derivative of above equation with respect to t. we get, da da - 2 [CE (1. et/Rc)] we know I da So, I [CE (1- it'lec)] I= ce Cl-etire ECE ['de et/re] I = CE[0- (R) atlec). Chloe () ze) I = CE (kciēt/R c) I h etike 1 I = To atteel... (1:15) OMI NOTE 7 PRO QUAL CAMERA