Exam 6: Capacitors and Inductors

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The voltage across an inductor with inductance of 510 mH is given by v(t)={0.05t2,0t<50.05(t10)2,5t<10 V0, otherwise v ( t ) = \left\{ \begin{array} { c } 0.05 t ^ { 2 } , \quad 0 \leq t < 5 \\0.05 ( t - 10 ) ^ { 2 } , \quad 5 \leq t < 10 \mathrm {~V} \\0 , \quad \text { otherwise }\end{array} \right. Find the current through the inductor i(t) and plot the voltage and current as a function of time.

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i(t)=1L0tv(λ)dλ={0.03268t3,0t<50.03268t30.9804t2+9.8039t24.5098,5t<10 A8.17,10ti ( t ) = \frac { 1 } { L } \int _ { 0 } ^ { t } v ( \lambda ) d \lambda = \left\{ \begin{array} { c } 0.03268 t ^ { 3 } , \quad 0 \leq t < 5 \\0.03268 t ^ { 3 } - 0.9804 t ^ { 2 } + 9.8039 t - 24.5098 , \quad 5 \leq t < 10 \mathrm {~A} \\8.17 , \quad 10 \leq t\end{array} \right.
i ( t ) = \frac { 1 } { L } \int _ { 0 } ^ { t } v ( \lambda ) d \lambda = \left\{ \begin{array} { c }  0.03268 t ^ { 3 } , \quad 0 \leq t < 5 \\ 0.03268 t ^ { 3 } - 0.9804 t ^ { 2 } + 9.8039 t - 24.5098 , \quad 5 \leq t < 10 \mathrm {~A} \\ 8.17 , \quad 10 \leq t \end{array} \right.
i ( t ) = \frac { 1 } { L } \int _ { 0 } ^ { t } v ( \lambda ) d \lambda = \left\{ \begin{array} { c }  0.03268 t ^ { 3 } , \quad 0 \leq t < 5 \\ 0.03268 t ^ { 3 } - 0.9804 t ^ { 2 } + 9.8039 t - 24.5098 , \quad 5 \leq t < 10 \mathrm {~A} \\ 8.17 , \quad 10 \leq t \end{array} \right.

voltage across a capacitor with capacitance of 39 is given by μF\mu \mathrm { F } v(t)={30t,0t<590t600,5t<1060t+900,10t<150, otherwise v ( t ) = \left\{ \begin{array} { c } - 30 t , \quad 0 \leq t < 5 \\90 t - 600 , \quad 5 \leq t < 10 \\- 60 t + 900 , \quad 10 \leq t < 15 \\0 , \quad \text { otherwise }\end{array} \right. Find the current through the capacitor i(t) and plot the voltage and current as a function of time.

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i(t)={1.17,0t<53.51,5t<102.34,10t<150, elsewhere mAi ( t ) = \left\{ \begin{array} { c } - 1.17 , \quad 0 \leq t < 5 \\3.51 , \quad 5 \leq t < 10 \\- 2.34 , \quad 10 \leq t < 15 \\0 , \text { elsewhere }\end{array} \quad \mathrm { mA } \right.    i ( t ) = \left\{ \begin{array} { c }  - 1.17 , \quad 0 \leq t < 5 \\ 3.51 , \quad 5 \leq t < 10 \\ - 2.34 , \quad 10 \leq t < 15 \\ 0 , \text { elsewhere } \end{array} \quad \mathrm { mA } \right.

Find the equivalent capacitance between terminals a and b of the circuit shown below. Find the equivalent capacitance between terminals a and b of the circuit shown below.

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La=L5 L6L7=27.92mHLb=L3+La=54.92mHLc=L2Lb=21.7457232mHLeq=L1+Lc=43.7457232mH\begin{aligned}\mathrm { L } _ { \mathrm { a } } & = \mathrm { L } _ { 5 } \left\| \mathrm {~L} _ { 6 } \right\| \mathrm { L } _ { 7 } = 27.92 \mathrm { mH } \\\mathrm { L } _ { b } & = \mathrm { L } _ { 3 } + \mathrm { L } _ { \mathrm { a } } = 54.92 \mathrm { mH } \\\mathrm { L } _ { \mathrm { c } } & = \mathrm { L } _ { 2 } \| \mathrm { L } _ { \mathrm { b } } = 21.7457232 \mathrm { mH } \\\mathrm { L } _ { e q } & = \mathrm { L } _ { 1 } + \mathrm { L } _ { \mathrm { c } } = 43.7457232 \mathrm { mH }\end{aligned} clear all; format long;
L1=22e3;L2=36e3;L3=27e3;L4=75e3;L5=87e3;L6=91e3;\mathrm { L } 1 = 22 \mathrm { e } - 3 ; \mathrm { L } 2 = 36 \mathrm { e } - 3 ; \mathrm { L } 3 = 27 \mathrm { e } - 3 ; \mathrm { L } 4 = 75 \mathrm { e } - 3 ; \mathrm { L } 5 = 87 \mathrm { e } - 3 ; \mathrm { L } 6 = 91 \mathrm { e } - 3 ;
La=P([L4, L5, L6])\mathrm { La } = \mathrm { P } ( [ \mathrm { L } 4 , \mathrm {~L} 5 , \mathrm {~L} 6 ] )
Lb=L3+La\mathrm { Lb } = \mathrm { L } 3 + \mathrm { La }
LcP([L2,Lb])\mathrm { Lc } - \mathrm { P } ( [ \mathrm { L } 2 , \mathrm { Lb } ] )
Leq=L1+Lc\mathrm { Leq } = \mathrm { L } 1 + \mathrm { Lc }

find the equivalent capacitance between terminals a and b of the circuit shown below. find the equivalent capacitance between terminals a and b of the circuit shown below.

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current through a capacitor with capacitance of 0.68 is given by i(t) = 4 μF\mu \mathrm { F } sin(2π×1500t)u(t)mA\sin ( 2 \pi \times 1500 t ) u ( t ) \mathrm { mA } . Find the voltage across the capacitor v(t)\mathrm { v } ( \mathrm { t } ) and plot the current and voltage as a function of time.

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Find the equivalent capacitance between terminals a and b of the circuit shown below. Find the equivalent capacitance between terminals a and b of the circuit shown below.

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 The current through a capacitor with capacitance of 33μF is given by \text { The current through a capacitor with capacitance of } 33 \mu \mathrm { F } \text { is given by } i(t)={0.005e1000(t+0.003),0t<0.0030.005e1000(t0.003),0.003t<0.006 A0, otherwise i ( t ) = \left\{ \begin{array} { c } 0.005 e ^ { - 1000 ( - t + 0.003 ) } , \quad 0 \leq t < 0.003 \\0.005 e ^ { - 1000 ( t - 0.003 ) } , \quad 0.003 \leq t < 0.006 \mathrm {~A} \\0 , \quad \text { otherwise }\end{array} \right.  Find the voltage across the capacitor v(t) and plot the current and voltage as a function of time. \text { Find the voltage across the capacitor } v ( t ) \text { and plot the current and voltage as a function of time. }

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the voltage across an inductor with inductance of 39 mH is given by v(t)=6cos(2π×3500t)v ( t ) = 6 \cos ( 2 \pi \times 3500 t ) u(t) V. Find the current through the inductor i(t) and plot the voltage and current as a function of time.

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the current through an inductor with inductance of 430 mH is given by i(t)={4t,0t<58t60,5t<104t+60,10t<150, otherwise Ai ( t ) = \left\{ \begin{array} { c } - 4 t , \quad 0 \leq t < 5 \\8 t - 60 , \quad 5 \leq t < 10 \\- 4 t + 60 , \quad 10 \leq t < 15 \\0 , \quad \text { otherwise }\end{array} \mathrm { A } \right. Find the voltage across the inductor v(t) and plot the current and voltage as a function of time.

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find the equivalent capacitance between terminals a and b of the circuit shown below. find the equivalent capacitance between terminals a and b of the circuit shown below.

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