Exam 7: RL and RC Circuits

The switch in the circuit shown below has been opened for a long time before it is closed at $t = 0$ . Find the current $i ( t )$ through the inductor for $t \geq 0$ . Assume $R _ { 1 } = 1 k \Omega , R _ { 2 } = 2 k \Omega , R _ { 3 } = 2$ $\mathrm { k } \Omega , \mathrm { R } _ { 4 } = 2 \mathrm { k } \Omega , \mathrm { I } _ { \mathrm { s } } = 5 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 6.4 \mathrm {~V} , \mathrm {~L} = 40 \mathrm { mH }$ . $The switch in the circuit shown below has been opened for a long time before it is closed at t = 0 . Find the current i ( t ) through the inductor for t \geq 0 . Assume R _ { 1 } = 1 k \Omega , R _ { 2 } = 2 k \Omega , R _ { 3 } = 2 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 2 \mathrm { k } \Omega , \mathrm { I } _ { \mathrm { s } } = 5 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 6.4 \mathrm {~V} , \mathrm {~L} = 40 \mathrm { mH } .$

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Question 1

Correct Answer:

Verified

$\mathrm { i } ( 0 ) = \mathrm { I } _ { 0 } = \mathrm { Is } \times \mathrm { R } _ { 1 } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { 2 } + \mathrm { R } _ { 3 } \right) = 1 \mathrm {~mA}$
From Is,
$\begin{array} { l } \mathrm { R } _ { \mathrm { b } } = \mathrm { R } _ { 2 } + \left( \mathrm { R } _ { 3 } \| \mathrm { R } _ { 4 } \right) = 3 \mathrm { k } \Omega \\\mathrm { I } _ { \mathrm { infl } } = \left[ \mathrm { I } _ { \mathrm { s } } \times \mathrm { R } _ { 1 } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { \mathrm { b } } \right) \right] \times \mathrm { R } _ { 3 } / \left( \mathrm { R } _ { 3 } + \mathrm { R } _ { 4 } \right) = 0.625 \mathrm {~mA}\end{array}$
From V $\mathrm { V } _ { \mathrm { s } }$
$\begin{array} { l } R _ { a } = R _ { 3 } \| \left( R _ { 1 } + R _ { 2 } \right) = 1.2 \mathrm { k } \Omega \\I _ { \text {inf } 2 } = \left[ V _ { \mathrm { s } } \times R _ { a } / \left( R _ { a } + R _ { 4 } \right) \right] / R _ { 3 } = 1.2 \mathrm {~mA} \\I _ { \text {inf } } = I _ { \text {infl } } + I _ { \text {inf } 2 } = 1.825 \mathrm {~mA} \\R _ { \text {eq } } = R _ { 3 } + \left[ R _ { 4 } \| \left( R _ { 2 } + R _ { 1 } \right) \right] = 3.2 \mathrm { k } \Omega \\\tau = L / R _ { \text {eq } } = 12.5 \mu \mathrm { s } , 1 / \tau = 80000 ( 1 / \mathrm { s } ) \\\mathrm { i } ( \mathrm { t } ) = \left[ \mathrm { I } _ { \text {inf } } + \left( \mathrm { I } _ { 0 } - \mathrm { I } _ { \text {inf } } \right) \mathrm { e } ^ { - 80000 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) = \left[ 1.825 + ( 1 - 1.825 ) \mathrm { e } ^ { - 80000 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) \\= \left[ 1.825 - 0.825 \mathrm { e } ^ { - 80000 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) \mathrm { mA }\end{array}$

switch 1 in the circuit shown below has been closed for a long time before it is opened at t = 0 and the switch 2 has been opened for a long time before it is closed at t = 0. Find the voltage $\mathrm { v } ( \mathrm { t } )$ across the capacitor for $\mathrm { t } \geq 0$ . Assume $\mathrm { R } _ { 1 } = 6 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 2 \mathrm { k } \Omega$ , $\mathrm { R } _ { 5 } = 1 \mathrm { k } \Omega , \mathrm { I } _ { \mathrm { s } } = 1 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 6 \mathrm {~V} , \mathrm { C } = 0.4 \mu \mathrm { F }$ . $switch 1 in the circuit shown below has been closed for a long time before it is opened at t = 0 and the switch 2 has been opened for a long time before it is closed at t = 0. Find the voltage \mathrm { v } ( \mathrm { t } ) across the capacitor for \mathrm { t } \geq 0 . Assume \mathrm { R } _ { 1 } = 6 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 2 \mathrm { k } \Omega , \mathrm { R } _ { 5 } = 1 \mathrm { k } \Omega , \mathrm { I } _ { \mathrm { s } } = 1 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 6 \mathrm {~V} , \mathrm { C } = 0.4 \mu \mathrm { F } .$

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Question 2

Correct Answer:

Verified

$\begin{array} { l } \mathrm { v } ( 0 ) = \mathrm { V } _ { 0 } = \left( \mathrm { R } _ { 3 } + \mathrm { R } _ { 4 } \right) \times \mathrm { I } _ { \mathrm { s } } \times \mathrm { R } _ { 1 } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { 2 } + \mathrm { R } _ { 3 } + \mathrm { R } _ { 4 } \right) = 1.5 \mathrm {~V} \\\mathrm { v } ( \infty ) = \mathrm { V } _ { \text {inf } } = \mathrm { V } _ { \mathrm { s } } \times \mathrm { R } _ { 4 } / \left( \mathrm { R } _ { 4 } + \mathrm { R } _ { 5 } \right) = 4 \mathrm {~V} \\\mathrm { R } _ { \mathrm { eq } } = \mathrm { R } _ { 3 } + \left( \mathrm { R } _ { 4 } \| \mathrm { R } _ { 5 } \right) = 1.6667 \mathrm { k } \Omega \\\tau = \mathrm { R } _ { \text {eq } } \mathrm { C } = 0.0006667 \mathrm {~s} , 1 / \tau = 1500 ( 1 / \mathrm { s } ) \\\mathrm { v } ( \mathrm { t } ) = \left[ \mathrm { V } _ { \text {inf } } + \left( \mathrm { V } _ { 0 } - \mathrm { V } _ { \text {inf } } \right) \mathrm { e } ^ { - 2500 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) = \left[ 4 + ( 1.5 - 4 ) \mathrm { e } ^ { - 1500 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) \\= \left[ 4 - 2.5 \mathrm { e } ^ { - 1500 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) \mathrm { V }\end{array}$

The switch in the circuit shown below has been opened for a long time before it is closed at $t = 0$ . Find the current $i ( t )$ through the inductor for $t \geq 0$ . Assume $R _ { 1 } = 1 k \Omega , R _ { 2 } = 1 k \Omega , R _ { 3 } = 2$ $\mathrm { k } \Omega , \mathrm { R } _ { 4 } = 2 \mathrm { k } \Omega , \mathrm { I } _ { \mathrm { s } } = 6 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 6 \mathrm {~V} , \mathrm {~L} = 50 \mathrm { mH }$ . $The switch in the circuit shown below has been opened for a long time before it is closed at t = 0 . Find the current i ( t ) through the inductor for t \geq 0 . Assume R _ { 1 } = 1 k \Omega , R _ { 2 } = 1 k \Omega , R _ { 3 } = 2 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 2 \mathrm { k } \Omega , \mathrm { I } _ { \mathrm { s } } = 6 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 6 \mathrm {~V} , \mathrm {~L} = 50 \mathrm { mH } .$

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Question 3

Correct Answer:

Verified

$\mathrm { R } _ { \mathrm { a } } = \left( \mathrm { R } _ { 2 } + \mathrm { R } _ { 1 } \right) \| \mathrm { R } _ { 3 } = 1 \mathrm { k } \Omega$
$\mathrm { i } ( 0 ) = \mathrm { I } _ { 0 } = \left[ \mathrm { V } _ { \mathrm { s } } \times \mathrm { R } _ { \mathrm { a } } / \left( \mathrm { R } _ { 4 } + \mathrm { R } _ { \mathrm { a } } \right) \right] / \mathrm { R } 3 = 1 \mathrm {~mA}$
From $\mathrm { V } _ { \mathrm { s } }$ ,
$\mathrm { I } _ { \text {infl } } = 1 \mathrm {~mA}$
From $\mathrm { I } _ { \mathrm { s } }$ ,
$\mathrm { R } _ { \mathrm { b } } = \mathrm { R } _ { 3 } \| \mathrm { R } _ { 4 } = 1 \mathrm { k } \Omega$
$\mathrm { R } _ { \mathrm { c } } = \mathrm { R } _ { 2 } + \mathrm { R } _ { \mathrm { b } } = 2 \mathrm { k } \Omega$
$\mathrm { I } _ { \mathrm { R } 2 } ( 0 ) = \mathrm { I } _ { \mathrm { S } } \times \mathrm { R } _ { 1 } / \left( \mathrm { R } _ { 1 } + \mathrm { R } _ { \mathrm { c } } \right) = 2 \mathrm {~mA}$
$I _ { \text {inf } 2 } = I _ { R 2 } ( 0 ) \times R _ { 4 } / \left( R _ { 3 } + R _ { 4 } \right) = 1 m A$
$\mathrm { I } _ { \mathrm { inf } } = \mathrm { I } _ { \mathrm { inf } 1 } + \mathrm { I } _ { \mathrm { inf } 2 } = 2 \mathrm {~mA}$
$R _ { \text {eq } } = R _ { 3 } + \left[ R _ { 4 } \| \left( R _ { 2 } + R _ { 1 } \right) \right] = 3 \mathrm { k } \Omega$
$\tau = \mathrm { L } / \mathrm { R } _ { \text {eq } } = 16.6667 \mu \mathrm { s } , 1 / \tau = 60000 ( 1 / \mathrm { s } )$
$\mathrm { i } ( \mathrm { t } ) = \left[ \mathrm { I } _ { \text {inf } } + \left( \mathrm { I } _ { 0 } - \mathrm { I } _ { \text {inf } } \right) \mathrm { e } ^ { - 60000 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) = \left[ 2 + ( 1 - 2 ) \mathrm { e } ^ { - 60000 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } )$ $= \left[ 2 - \mathrm { e } ^ { - 60000 \mathrm { t } } \right] \mathrm { u } ( \mathrm { t } ) \mathrm { mA }$