Calculate Impedance Z

To find the nature of the circuit, we calculate the impedance $Z = \frac{v}{i}$. Given $v = 10 + j5$ and $i = 6 + j4$.

$Z = \frac{10 + j5}{6 + j4}$

Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator, $(6 - j4)$.

$Z = \frac{(10 + j5)(6 - j4)}{(6 + j4)(6 - j4)} = \frac{60 - j40 + j30 + 20}{36 + 16} = \frac{80 - j10}{52}$

Determine reactance sign

Simplifying the result gives $Z = \frac{80}{52} - j\frac{10}{52} = 1.538 - j0.192 \Omega$. The imaginary part of $Z$ (reactance $X$) is negative.

$X = -0.192 \Omega$

Conclusion

Since the imaginary part of the impedance $X$ is negative ($X < 0$), the circuit is capacitive.

$X < 0 \implies \text{Capacitive}$