The alternating current is given by

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Question

${f}_{{rms}}^{2}={f}_{1 {m} {m}}^{2}+{f}_{2 {rms}}^{2}$

$=\left(\frac{\sqrt{42}}{\sqrt{2}}\right)^{2}+10^{2}$

$=121 \Rightarrow {f}_{{rms}}=11\,

Vedclass · Accepted Answer

$11$

Vedclass · Answer

${f}_{{rms}}^{2}={f}_{1 {m} {m}}^{2}+{f}_{2 {rms}}^{2}$

$=\left(\frac{\sqrt{42}}{\sqrt{2}}\right)^{2}+10^{2}$

$=121 \Rightarrow {f}_{{rms}}=11\, {A}$

The alternating current is given by

${i}=\left\{\sqrt{42} \sin \left(\frac{2 \pi}{{T}} {t}\right)+10\right\} {A}$

The $r.m.s.$ value of this current is ${A}$

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The alternating current is given by ${i}=\left\{\sqrt{42} \sin \left(\frac{2 \pi}{{T}} {t}\right)+10\right\} {A}$ The $r.m.s.$ value of this current is ${A}$