$X_{L}$ is $I=\frac{E}{R+\left(X_{L}-X_{C}\right)}$. . Use this law to solve the problem.
-Find $X_{C}$ if $I=6-2 i, E=5-5 i, R=9$, and $X_{L}=1$.
A) $\frac{3}{2}+8 \mathrm{i}$
B) $\frac{3}{2}-8 \mathrm{i}$
C) $-\frac{3}{2}-8 \mathrm{i}$
D) $-\frac{3}{2}+8 \mathrm{i}$

Question

$X_{L}$ is $I=\frac{E}{R+\left(X_{L}-X_{C}\right)}$. . Use this law to solve the problem.
-Find $X_{C}$ if $I=6-2 i, E=5-5 i, R=9$, and $X_{L}=1$.
A) $\frac{3}{2}+8 \mathrm{i}$ 
B) $\frac{3}{2}-8 \mathrm{i}$ 
C) $-\frac{3}{2}-8 \mathrm{i}$ 
D) $-\frac{3}{2}+8 \mathrm{i}$

Quizplus · Accepted Answer

Given $I=6-2i$, $E=5-5i$, and $R=9$, we can find $X_C$ using Ohm's law and the impedance formula for a series circuit. The total impedance $Z = R + jX_L + jX_C$, where $j$ is the imaginary unit. We know $Z = \frac{E}{I}$, so we can solve for $X_C$. Plugging in the given values, we find $X_C = \frac{3}{2} - 8i$, which matches option B.

XLX_{L}XL​ Is I=ER+(XL−XC)I=\frac{E}{R+\left(X_{L}-X_{C}\right)}I=R+(XL​−XC​)E​ Use This Law to Solve the Problem XCX_{C}XC​

$X_{L}$ Is $I=\frac{E}{R+\left(X_{L}-X_{C}\right)}$ Use This Law to Solve the Problem $X_{C}$