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Question Number 29209 by tawa tawa last updated on 05/Feb/18

Commented by tawa tawa last updated on 05/Feb/18

$please help with . 4, 5, 6, 7$
Commented by ajfour last updated on 05/Feb/18

Commented by ajfour last updated on 05/Feb/18

$Q . 4)$ $l e t c h a r g e a t t i m e t b e Q .$ $A n d t h e i n i t i a l c h a r g e Q_{0} .$ ${K i r c h o f f}^{'} s v o l t a g e l a w g i v e s$ $\frac{Q}{C} - i R = 0$ $\Rightarrow \frac{Q}{R C} = - i = \frac{d Q}{d t}$ $\Rightarrow \int_{Q_{0}}^{Q} \frac{d Q}{Q} = - \int_{0}^{t} \frac{d t}{R C}$ $o r \ln \frac{Q}{Q_{0}} = - \frac{t}{R C}$ $\Rightarrow Q = Q_{0} e^{- \frac{t}{R C}} .$ $Q . 5)$ $v_{0} = \frac{Q_{0}}{C}, v = \frac{Q}{C}$ $S o f r o m r e s u l t o f Q . (4)$ $v = v_{0} e^{- \frac{t}{R C}} .$
Commented by tawa tawa last updated on 05/Feb/18

$God bless you sir . i really appreciate$
	Answered by ajfour last updated on 05/Feb/18
	$Q.6) ((Ldi)/dt)+iR = Ecos ωt ⇒ (di/dt) +((iR)/L) = (E/L)cos ωt integrating factor for the linear differential equation is e^(∫(R/L)dt) =e^(Rt/L) so i(e^(Rt/L) )=(E/L)∫e^(Rt/L) cos ωt =(E/L)[(((sin ωt)/ω))e^(Rt/L) −(R/(ωL))∫e^(Rt/L) sin ωtdt ]+c =(E/L){((e^(Rt/L) sin ωt)/ω)−(R/(ωL))[−((e^(Rt/L) cos ωt)/ω)+(R/(ωL))∫e^(Rt/L) cos ωtdt ]}+c_1 =(E/L){((e^(Rt/L) sin ωt)/ω)−(R/(ωL))[−((e^(Rt/L) cos ωt)/ω)+(R/(ωL))×(L/E)(ie^(Rt/L) )]}+c_1 ⇒ i(e^(Rt/L) )[1+(R^2 /(ω^2 L^2 ))]=(E/(ωL))(e^(Rt/L) )[sin ωt+(R/(ωL))cos ωt]+c_1 ⇒ i = (E/(𝛚^2 L^2 +R^2 ))(𝛚Lsin 𝛚t+Rcos 𝛚t)+c_2 e^(−Rt/L) if i=0 at t=0 0=((ER)/(ω^2 L^2 +R^2 ))+c_2 ⇒ i=(E/(ω^2 L^2 +R^2 ))(ωLsin ωt+Rcos ωt−Re^(−Rt/L) ) .$
	$Q . 6) \frac{L d i}{d t} + i R = E \cos ω t$ $\Rightarrow \frac{d i}{d t} + \frac{i R}{L} = \frac{E}{L} \cos ω t$ $i n t e g r a t i n g f a c t o r f o r t h e l i n e a r$ $d i f f e r e n t i a l e q u a t i o n i s$ $e^{\int \frac{R}{L} d t} = e^{R t / L}$ $s o i (e^{R t / L}) = \frac{E}{L} \int e^{R t / L} \cos ω t$ $= \frac{E}{L} [(\frac{\sin ω t}{ω}) e^{R t / L} - \frac{R}{ω L} \int e^{R t / L} \sin ω t d t] + c$ $= \frac{E}{L} {\frac{e^{R t / L} \sin ω t}{ω} - \frac{R}{ω L} [- \frac{e^{R t / L} \cos ω t}{ω} + \frac{R}{ω L} \int e^{R t / L} \cos ω t d t]} + c_{1}$ $= \frac{E}{L} {\frac{e^{R t / L} \sin ω t}{ω} - \frac{R}{ω L} [- \frac{e^{R t / L} \cos ω t}{ω} + \frac{R}{ω L} \times \frac{L}{E} ({i e}^{R t / L})]} + c_{1}$ $\Rightarrow i (e^{R t / L}) [1 + \frac{R^{2}}{ω^{2} L^{2}}] = \frac{E}{ω L} (e^{R t / L}) [\sin ω t + \frac{R}{ω L} \cos ω t] + c_{1}$ $\Rightarrow i = \frac{E}{ω^{2} L^{2} + R^{2}} (ω L \sin ω t + R \cos ω t) + c_{2} e^{- R t / L}$ $i f i = 0 a t t = 0$ $0 = \frac{E R}{ω^{2} L^{2} + R^{2}} + c_{2}$ $\Rightarrow i = \frac{E}{ω^{2} L^{2} + R^{2}} (ω L \sin ω t + R \cos ω t - {R e}^{- R t / L}) .$
	Commented by tawa tawa last updated on 05/Feb/18

	$wow, God bless you sir .$
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