Answer
(a) $0.173V$
(b) $7.84V$
(c) $1.84V$
(d) greater than
Work Step by Step
(a) We know that
$V_{rms, R}=I_{rms}\times R$
$\implies V_{rms, R}=0.069\times 25$
$V_{rms,R}=0.173V$
(b) The voltage across inductor L is given as
$V_{rms, L}=I_{rms}\times X_L$
We plug in the known values to obtain:
$V_{rms, L}=0.069\times 113.14$
$V_{rms, L}=7.84V$
(c) The voltage across capacitor is given as
$V_{rms, C}=I_{rms}\times X_C$
We plug in the known values to obtain:
$V_{rms, C}=0.069\times 26.5$
$V_{rms, C}=1.84V$
(d) We know that the sum of the rms voltages in parts (a), b and (c) is greater than 6V.
As $V_{rms}=I_{rms}\times Z$
but $Z=\sqrt{R^2+(X_L-X_C)^2}$
and $Z\ne R+X_L+X_C$
Therefore $V_{rms}\ne V_{rms, R}+V_{rms,C}+V_{rms,L}$
We can see that $V_{rms, R}+V_{rms,C}+V{rms, L}\gt V_{rms}$
Thus, the sum of magnitudes of the voltages that are out of phase is greater than the magnitude of their vector sum.
