Answer & Solution
Answer: Option B
Solution:
$$\eqalign{
& {{\text{E}}_{{\text{rms}}}} = \sqrt {{\text{V}}_{\text{L}}^2} + {\text{V}}_{\text{R}}^2 \cr
& \Rightarrow {{\text{E}}_{{\text{rms}}}} = \sqrt {{{12}^2}} + {14^2} = \sqrt {340{\text{V}}} \cr
& {{\text{E}}_{{\text{max}}}} = {{\text{E}}_{{\text{rms}}}}\sqrt 2 \cr
& {{\text{E}}_{{\text{max}}}} = \sqrt {680{\text{V}}} \cr
& \Rightarrow {{\text{E}}_{{\text{max}}}} = 26.0\,{\text{V}} \cr} $$