Question-29209

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) \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 .

wow, God bless you sir .