No got the solution of the Equation (2) (0)(C)/10 + 1/10 (f) (₁0.1 +0.1 shere the integration constant Ny can find the v

[answerhappygod]

No Got The Solution Of The Equation 2 0 C 10 1 10 F 0 1 0 1 Shere The Integration Constant Ny Can Find The V 1 (32.69 KiB) Viewed 28 times

No got the solution of the Equation (2) (0)(C)/10 + 1/10 (f) (₁0.1 +0.1 shere the integration constant Ny can find the value of integration constant, Cng the initial condition w=0, (0) (0) 0 Then hing the given inal condition, from Equation (3) () (C₁0.1 +0.1 (0) (C₁)0.1 +0.1 (0) (C₁)0.1² +0.1|| (0)(Cx 0.1×1) +01) +0=0.IC+0.1 BIC)-(-0.1) +9= =5-50 Therefore pluggedin (-1) Eation no g Về ANH R (t)=(-1)×01 x) +01) (f)-0.1-(0.1) Ne finally got the Solution (0011- ANSWER: dow om the defination of the Zamientcament, we already know fransient current is an oscillatory or periodic current that flows in a circuit short finn lectromagnetic distuntance; / ind When the current at each point in the circult is constant does not change with mettermed as the deady state current The Transient current i The SteadyState currenti "\"/ ⇒ i(t) = lim (0.1) - lim (0.1e-50) →i(t) = (0.1) (0.1e-50x∞) ⇒ i(t) = [(0.1) - = [(0.1)-(S)] ⇒ i(t) = [(0.1) – (-¹)] ⇒ i(t) = [(0.1) - 0] ⇒ i(t) = 0.1 Amp. So, The Transient current is: i(t) = 0.1(1-e-50) Amp. [ANSWER-2] And The Steady State current is: i(t) = 0.1 Amp. [ANSWER-3]