Question

A coil having self-inductance $L$ is connected in series with a bulb $\text{B}$ with an AC source. Brightness of the bulb decrease when-

• A. An iron rod is inserted in the coil
• B. Frequency of the AC source is decreased
• C. Number of turns in the coil is reduced
• D. A capacitance of reactance $X_C=X_L$ included in the same circuit.

Answer 1

The correct option is A An iron rod is inserted in the coil

The initial active power in the circuit is given by,
$P=I_{rms}^{2}R={\left(\frac{E_{rms}}{Z}\right)}^{2}R$

Here, $Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(\omega L{)}^2}$

$\therefore P=\left(\frac{(E_{rms}{)}^2}{R^2+(\omega L{)}^2}\right)R$

If the frequency of the AC supply is decreased, $Z$ decreases so that the power, and hence, the brightness of the bulb will increase.

If the number of turns in the coil are decreased, its self inductance $L$ will decrease. This will lead to a decrease in $Z$ so that the power, and hence, the brightness of the bulb will increase.

When a capacitance of reactance $X_C=X_L$ is connected in the given circuit, the current and emf gets in phase so that $R=Z$. As a result of this, the the power, and hence, the brightness of the bulb will increase $\left(\text{As}\right(Z>R)\text{when}\varphi >0$.

When an iron rod is inserted in the coil, the self inductance $L$ of the coil increases so that $Z$ increases.
According to the equation of power, the power decreases and therefore, brightness of the bulb also decreases.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.