Question

When $100V$ dc is applied across a solenoid, a current of $1.0A$ flows in it. When $100V$ac is applied across the same coil, the current drops to $0.5A$. If the frequency of the ac source is $50Hz$, the impedance and inductance of the solenoid are
A. $200\Omega$ and $0.55H$
B. $100\Omega$ and $0.86H$
C. $200\Omega$ and $1.0H$
D. $200\Omega$ and $0.93H$

Answer 1

We have to know the definition of dc current and the definition of ac current. Here dc current means direct current. Which flows in one direction for example the current that flows in a flashlight running on batteries is called dc current . Now the ac current means the alternating currents. Which periodically reverses direction and changes its magnitude constantly.

Complete step by step answer:
Voltage of dc is $100V$, current of dc is $1A$.
${V_{dc}} = 100V \\\ \Rightarrow{I_{dc}} = 1A$
Now, ${V_{rms}} = 100V$ this is also given in the question. And also another given information is,
$f = 50Hz$ ,
$\Rightarrow{I_{rms}} = 0.5A$
In dc, the resistance of the solenoid by Ohm’s Law is equal to,
$\dfrac{{{V_{dc}}}}{{{I_{dc}}}} = \dfrac{{100}}{1}\Omega$
Now, in ac impedance z is equal to,

\Rightarrow\dfrac{{{V_{rms}}}}{{{I_{rms}}}}= 200\Omega $$ Now, solenoid is an inductor. So in the second case, this became a L-R circuit. Impedance is equal to $$\sqrt {{R^2} + {X_L}^2} $$ $200 = \sqrt {({{100}^2} + {L^2}{\omega ^2})} \\\ \Rightarrow 40000 = 10000 + {L^2}{(2\pi \times 50)^2} $ $$\Rightarrow30000 = {L^2}{(2\pi \times 50)^2}$$ $$\Rightarrow \sqrt 3 = L(2\pi \times 50)$$ $$\Rightarrow L = \dfrac{{100 \times \sqrt 3 }}{{2\pi \times 50}}$$ $$\therefore L = 0.55H$$ So the inductance is $$0.55H$$ **So the right option would be option number A.** **Note:** We can get confused between the dc current, voltage and the ac rms current and rms voltages. To avoid this type of confusion we should briefly know what is rms voltage and rms current. We applied Ohm’s Law here, so we have to know the law properly.