Question

In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?

• A. Only a capacitor
• B. Resistor and capacitor
• C. Only resistor
• D. Resistor and an inductor

Answer 1

Answer: (A), (B)

Resistor and capacitor

Formula used:

$X_L=2\pi vL$ ,$X_C=\frac{1}{2\pi vC,}$

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

$I_{rms}=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$I_{rms}=\frac{V_{rms}}{\sqrt{R^2}}$

Suppose circuit contain only resistor, then R does not depend on frequency it means current should have remained same which is not possible.

$I_{rms}=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$I_{rms}=\frac{V_{rms}}{\sqrt{R^2+(X_L{)}^2}}$

$X_L=2\pi \nu L$

Suppose if we increase frequency, then $X_L$ also increases and thus impedance will increase it means current will decrease. This circuit is not possible according to given condition.
Again,

$I_{rms}=\frac{V_{rms}}{\sqrt{R^2+(X_C{)}^2}}$

$\therefore X_C=\frac{1}{2\pi vC}$

As we increase frequency reactance of capacitance will decrease and impedance decreases. It implies $I_{rms}$ value of current increase so, this element is possible in circuit.

The reactance of a capacitor is

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi vC}$

Therefore, likely elements constituting the circuit may be capacitor only or capacitor and a resistor.

Final Answer: Option (c), (d)