91Ó°ÊÓ

• Maximum generator emf,$n_m=125\mathrm{V}$.
• Maximum current,$I_m=3.20\mathrm{A}$.
• Currently leads the generator emf by $\mathrm{Ï•}=0.982\mathrm{rad}$.

The resistance in an AC circuit is called impedance. The resistance due to the capacitor in an AC circuit is called capacitive reactance and the resistance due to the inductor is called inductive reactance.

Use the formula of impendence to determine the impendence of the circuit. Then, using the formula of Ohm’s law, determine the resistance of the circuit. As the current lead the generator emf, using this given information, find out whether the circuit is predominantly capacitive or inductive.

The formulae are as follows:

$Z=n_m/I$.

Here, ${\mathrm{n}}_{\mathrm{m}}$is EMF amplitude voltage, I is current.

$V_R=IR=n_m\cos \mathrm{Ï•}$.

Here, $V_R$is the voltage across the resistor, I is current, ${\mathrm{n}}_{\mathrm{m}}$is EMF amplitude voltage, and is$\mathrm{Ï•}$ the phase angle.

The impendence of the circuit is given by,

$Z=\frac{n_m}{I}$.

Substituting the given values,

$\begin{array}{c}Z=\frac{125\mathrm{V}}{3.20\mathrm{A}}\\ Z=39.1\mathrm{Î\copyright }\end{array}$

Hence, the impendence of the circuit is $39.1\mathrm{Î\copyright }$.

The given current leads the emf by.

Since,${\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}=\sin \mathrm{Ï•}=\sin (0.982\mathrm{rad})$.

So,$X_L<X_C$.

Hence, the circuit is predominantly capacitive.

Find the impendence of the circuit by using the formula of impendence of the circuit. Find the resistance of the circuit by using Ohm’s law. Find that the circuit is capacitive.