Question

Alternating current is flowing in inductance L and resistance R. The frequency of source is $\omega /2\pi$which of the following statements:
A. For low frequency the limiting value of impedance is L.
B. For high frequency the limiting value of impedance is $\omega L$
C. For high frequency the limiting value of impedance is R.
D. For low frequency the limiting value of impedance is $\omega L$

Answer 1

To solve this question, we have to know about impedances. We know that, the all-out resistance to substituting flow by an electric circuit, equivalent to the square base of the amount of the squares of the obstruction and reactance of the circuit and normally communicated in ohms. Image: Z. this is called impedance.

Complete step by step answer:
To solve this question we have to calculate the impedance first,
$Z = \sqrt {{{({X_c} - {X_l})}^2} + {R^2}}$
Here Z is the impedance and R is the resistance.
Now, here in this question the capacitor is not there, so ${X_c}$ is zero. So,
$Z = \sqrt {{X_L}^2 + {R^2}}$
Or, $Z = \sqrt {{\omega ^2}{l^2} + {R^2}}$
For low frequency we know that,
$\omega < < l$
Now, again, from the above equation, we get that, v
$Z = \sqrt {{\omega ^2}{l^2} + {R^2}}$
Which is equal to,
\sqrt {{R^2}(1 + \dfrac{{{\omega ^2}{l^2}}}{{{R^2}}})} = R\sqrt {(1 + \dfrac{{{\omega ^2}{l^2}}}{{{R^2}}})} \\\
For high frequency we can say,
$Z = \sqrt {{\omega ^2}{l^2}(1 + \dfrac{{{R^2}}}{{{\omega ^2}{l^2}}})}$
Here, for high frequency, $\dfrac{{{R^2}}}{{{\omega ^2}{l^2}}} < < 1$
That means, $Z = \omega l$
From this we can say, for high frequency the limiting value of impedance is $\omega l$.

So, the right option is B.

Note: We also have to know that, the trademark impedance (normally composed Z0) of a uniform transmission line is the proportion of the amplitudes of voltage and current of a solitary wave proliferating along the line; that is, a wave going one way without appearance the other way.