An electric bulb,an open coil inductor,an AC | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) When an iron rod is inserted into the inductor,the self-inductance $L$ of the coil increases because the permeability of the core increases. The i

Vedclass · Accepted Answer

The glow of the bulb decreases

Vedclass · Answer

(C) When an iron rod is inserted into the inductor,the self-inductance $L$ of the coil increases because the permeability of the core increases. The inductive reactance of the circuit is given by $X_L = \omega L$. As $L$ increases,$X_L$ increases. The total impedance of the series circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb. Since $X_L$ increases,the total impedance $Z$ of the circuit increases. The current in the circuit is given by $I = V/Z$. As $Z$ increases,the current $I$ flowing through the bulb decreases. Since the power dissipated in the bulb is $P = I^2 R$,a decrease in current leads to a decrease in the power,and thus the glow of the bulb decreases.

An electric bulb,an open coil inductor,an $AC$ source,and a key are all connected in series to form a closed circuit. The key is closed and after some time,an iron rod is inserted into the interior of the inductor. Then:

Similar Questions

Calculate the capacitive reactance of a capacitor in order to run a bulb rated at $10 \, W, 60 \, V$ when connected in series to an $a.c.$ source of $100 \, V$.

The power factor of an $R-L$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of $AC$ is doubled,the power factor will now be

Alternating current is flowing in an $LR$ circuit with inductance $L$ and resistance $R$. The frequency of the source is $f = \omega / 2\pi$. Which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module