Question

Alternating emf of $e = 220\sin 100\pi t$ is applied to a circuit containing an inductance of $\dfrac{1}{\pi }$ henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected to the circuit?

Answer 1

First, the general equation for the alternating emf has to be concluded. From the equation, the peak value of the alternating current can be found. Note that, The equation for instantaneous current through the circuit can be written by comparing the given alternating emf with the general equation for alternating emf and then calculate the peak value of alternating current.

Formula used:
The general equation of alternating emf is $e = {e_0}\sin \omega t$.
The value of peak value of alternating current ${i_0} = \dfrac{{{e_0}}}{{\omega L}}$
The general equation of instantaneous current is $i = {i_0}\sin \left( {\omega t - \dfrac{\pi }{2}} \right)$.

Complete answer:
Given, alternating emf is $e = 220\sin 100\pi t$.
Given, the value of inductance $L$ is $\dfrac{1}{\pi }{\text{ H}}$
The general equation of alternating emf is given by the equation $e = {e_0}\sin \omega t$.
Compare the given alternating current with the general equation.
So, ${e_0} = 220$, and $\omega = 100\pi$.
Calculate the value of peak value of alternating current.
${i_0} = \dfrac{{{e_0}}}{{\omega L}}$
Substitute $220$ for ${e_0}$, $100\pi$for $\omega$, and $\dfrac{1}{\pi }$ for $L$ in the above equation.
${i_0} = \dfrac{{220}}{{\left( {100\pi } \right)\left( {\dfrac{1}{\pi }} \right)}}$
$\Rightarrow {i_0} = \dfrac{{220}}{{100}}$
$\Rightarrow {i_0} = 2.2{\text{ A}}$
The general equation of instantaneous current is given by the equation $i = {i_0}\sin \left( {\omega t - \dfrac{\pi }{2}} \right)$.
Substitute $2.2$ for ${i_0}$, and $100\pi$for $\omega$ in the above equation.
$i = 2.2\sin \left( {100\pi t - \dfrac{\pi }{2}} \right)$
Therefore, the equation for instantaneous current through the circuit is $i = 2.2\sin \left( {100\pi t - \dfrac{\pi }{2}} \right)$ in Amperes.
The reading of the Galvanometer can be calculated from the RMS value of current flowing through the circuit which is given as follows.
${i_{rms}} = \dfrac{{{i_0}}}{{\sqrt 2 }}$
Substitute $2.2$ for ${i_0}$in the above equation.
${i_{rms}} = \dfrac{{2.2}}{{\sqrt 2 }}$
$\Rightarrow {i_{rms}} = \dfrac{{2.2}}{{1.41}}$
$\Rightarrow {i_{rms}} = 1.56$
So, the RMS value of current flowing through the circuit is $1.56$ in Amperes.
Therefore, the reading of the Galvanometer is $1.56{\text{ A}}$

Note:
The emf or voltage whose value fluctuates sinusoidally concerning time is known as alternating emf. The corresponding current is known as alternating current.
The instantaneous current is defined as the current flowing through the circuit at a given instant of time.
The root means square or RMS value of current is the average value of current in the circuit.