Question

For a certain driven series RLC circuit, the maximum generator emf is 125V and the maximum current is 3.20A . If the current leads the generator emf by 0.982rad , what are the
(a) impedance and (b) resistance of the circuit? (c) is the circuit predominantly capacitive or inductive?

Answer 1

Hint : To solve this question, first we will rewrite the given facts or data of the question. And then we will first find the impedance of the circuit by applying its formula. Similarly, we can find the resistance by applying a resistance formula.

Complete Step By Step Answer:
Given that-
The maximum generator emf is $125V$ .
The maximum current is $3.20A$ .
(a). So, the impedance is: (it is denoted by $Z$ )
$Z = \dfrac{{{E_m}}}{I} = \dfrac{{125V}}{{3.20A}} = 39.1\Omega$ .
Here, ${E_m}$ is the maximum emf of the generator.
and, $I$ is the maximum current.
(b). To find the resistance of the circuit:
We will relate the given facts as follows:
From ${V_R} = I.R = {E_m}.\cos \phi$ , we get:
$R = \dfrac{{{E_m}.\cos \phi }}{I}$
As we have the value of $\phi$ is $0.982rad$ .
$\therefore R = \dfrac{{(125V)\cos (0.982rad)}}{{3.20A}} = 21.7\Omega$
(c). Since, ${X_L} - {X_C}\alpha \sin \phi = \sin ( - 0.982rad)$
we conclude that ${X_L} < {X_C}$ .
Hence, the circuit is predominantly capacitive.

Note :
Resistance is a concept used for DC (direct currents) whereas impedance is the AC (alternating current) equivalent. Resistance is due to electrons in a conductor colliding with the ionic lattice of the conductor meaning that electrical energy is converted into heat.