Two elements P and Q are connected in series as shown in the figu

SOLUTION

Concept:

For a series circuit, the net impedance is given by:

\(z = \frac{V}{I} = {R_{net}} + j\;{X_{net}}\)

R_net = Net resistance for the series circuit

X_net = Net reactance.

Also, if V(t) = V_m(cos ωt + θ)

The phasor representation is given by:

V = V_m ∠θ

Application:

For the given circuit:

V_s (supply voltage) = 24 cos (4t + 30°), i.e.

V_s = 24 ∠30°

Similarly I (Circuit current) = 0.5 cos (4t – 30°) A, i.e.

I = 0.5 ∠ -30°

The net impedance is defined as the ratio of the supply voltage and the circuit current, i.e.

\(\frac{{{V_s}}}{I} = \frac{{24\angle 30^\circ }}{{0.5\;\angle - 30^\circ }}\)

\(\frac{{{V_s}}}{I} = 48\;\angle 30^\circ - \left( { - 30^\circ } \right)\)

\(\frac{{{V_s}}}{I} = 48\;\angle 60^\circ\)

\(\frac{{{V_s}}}{I} = Z = 48\cos 60^\circ + 48\;j\sin 60^\circ\)

Z = 24 + 24√3 j

Since, Z = R_net + j X_net, we conclude that:

Element P must be a resistor of 24 Ω.

Also, since the voltage leads the current, we conclude that the element Q must be an inductor with:

X_L = 24√2, i.e.

ω L = 24√3

With ω = 4, the above equation can be written as:

4L = 24√3

L = 6√3 = 10.4 H

Note: If the net impedance had a negative imaginary part, the element Q would have been a capacitance.