Question

An e.m.f. \[E = 4\cos \left( {1000t} \right)\,{\text{volt}}\] is applied to an LR circuit of inductance \[3\,{\text{mH}}\] and resistance \[4\,{\text{ohms}}\]. The amplitude of current in the circuit is:
A.\[\dfrac{4}{{\sqrt 7 }}\,{\text{A}}\]
B.\[1.0\,{\text{A}}\]
C.\[\dfrac{4}{7}\,{\text{A}}\]
D.\[0.8\,{\text{A}}\]

Answer 1

Hint:Use the formula for the impedance of the given circuit. Also use the Ohm’s law. Use Ohm’s law in the form of the impedance of the circuit to determine the amplitude of current in the given circuit.

Formulae used:
The formula for the impedance is
\[Z = \sqrt {{R^2} + {\omega ^2}{L^2}} \] …… (1)
Here, \[Z\] is the impedance, \[R\] is the resistance, \[L\] is the inductance and \[\omega \] is the angular frequency.
The equation for Ohm’s law is
\[I = \dfrac{V}{R}\] …… (2)
Here, \[I\] is the current, \[V\] is the potential difference and \[R\]is the resistance.

Complete step by step answer:
We see that the equation for the emf applied to the LR circuit is \[E = 4\cos \left( {1000t} \right)\,{\text{volt}}\]. The inductance is \[3\,{\text{mH}}\] and the resistance is \[4\,\Omega \]. We know that the equation of emf of a circuit is \[E = {E_0}\cos \omega t\]. Let us compare the given equation for emf with the equation \[E = {E_0}\cos \omega t\] to determine the angular frequency and potential difference. From this comparison, the angular frequency \[\omega \] is \[1000\,{\text{rad/s}}\] and the potential difference in the circuit is \[4\,{\text{V}}\].

Let us now calculate the impedance of the LR circuit.
Substitute \[4\,\Omega \] for \[R\], \[1000\,{\text{rad/s}}\] for \[\omega \] and \[3\,{\text{mH}}\] for \[L\] in equation (1).
\[Z = \sqrt {{{\left( {4\,\Omega } \right)}^2} + {{\left( {1000\,{\text{rad/s}}} \right)}^2}{{\left( {3\,{\text{mH}}} \right)}^2}} \]
\[ \Rightarrow Z = \sqrt {{{\left( {4\,\Omega } \right)}^2} + {{\left( {1000\,{\text{rad/s}}} \right)}^2}{{\left[ {\left( {3\,{\text{mH}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{H}}}}{{1\,{\text{mH}}}}} \right)} \right]}^2}} \]
\[ \Rightarrow Z = \sqrt {16 + 9} \]
\[ \Rightarrow Z = \sqrt {25} \]
\[ \Rightarrow Z = 5\,\Omega \]
Hence, the impedance of the given LR circuit is \[5\,\Omega \].

We should now calculate the current in the LR circuit.
We can rewrite Ohm’s law using the impedance instead of resistance of the LR circuit.
\[I = \dfrac{V}{Z}\]
Substitute \[4\,{\text{V}}\] for \[V\] and \[5\,\Omega \] for \[Z\] in the above equation.
\[I = \dfrac{{4\,{\text{V}}}}{{5\,\Omega }}\]
\[ \therefore I = 0.8\,{\text{A}}\]

Therefore, the amplitude of current in the given LR circuit is \[0.8\,{\text{A}}\]. Hence, the correct option is D.

Note: The quantity amplitude of the electric current of the circuit is the same as that of the normal electric current. The emf of the circuit may be given in terms of the phase difference. In such cases, one should use the standard equation of emf in terms of phase difference to compare and determine the given quantities in the equation.