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A 10 H inductor, a 40 \mu F capacitor, and a voltage supply whose voltage is given by 100sin(50t)...

Question:

A {eq}10 H{/eq} inductor, a {eq}40 \mu F{/eq} capacitor, and a voltage supply whose voltage is given by {eq}100sin(50t){/eq} are connected in series in an electrical circuit. Find the current as a function of the time if the initial charge on the capacitor is zero and the initial current is zero.

Electric Current:

An electrical quantity that represents the movement of electric charges with time is known as electric current. Its measurable unit is Ampere. It moves in the opposite direction of movement of electron.

Answer and Explanation:


Given Data:

  • The inductance of inductor is: {eq}{L_o} = 10\;{\rm{H}} {/eq}
  • The capacitance of capacitor is: {eq}{C_o} = 40\;\mu {\rm{F}} {/eq}
  • The supply voltage is: {eq}{V_o} = 100\sin \left( {50t} \right) {/eq}


The angular frequency is: {eq}{\omega _o} = 50\;{\rm Hz} {/eq}


The expression for inductor reactance is

{eq}{X_L} = {\omega _o}{L_o} {/eq}


The expression for capacitor reactance is

{eq}{X_C} = \dfrac{1}{{{\omega _o}{C_o}}} {/eq}


The expression for total impedance of the circuit is

{eq}{Z_o} = j{X_L} - j{X_C} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} {Z_o} &= j{\omega _o}{L_o} - j\dfrac{1}{{{\omega _o}{C_o}}}\\ &= j\left[ {50 \times 10 - \dfrac{1}{{50 \times 40 \times {{10}^{ - 6}}}}} \right]\\ &= j\left[ {500 - \dfrac{{1000000}}{{2000}}} \right]\\ &= j\left[ {500 - 500} \right]\\ &= 0 \end{align*} {/eq}


The expression for current in circuit is

{eq}{I_o} = \dfrac{{{V_o}}}{{{Z_o}}} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} {I_o} &= \dfrac{{100\sin \left( {50t} \right)}}{0}\\ &= \infty \end{align*} {/eq}


Thus the current in circuit is {eq}\infty {/eq} that means it short circuit


Learn more about this topic:

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Change in Electric Current: Physics Lab

from Physics: High School

Chapter 16 / Lesson 2
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