JEE Mains · Physics · STD 12 -7. Alternating current
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 \({A}\)
- A \(11\)
- B \(13\)
- C \(9\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(11\)
Step-by-step Solution
Detailed explanation
\({f}_{{rms}}^{2}={f}_{1 {m} {m}}^{2}+{f}_{2 {rms}}^{2}\) \(=\left(\frac{\sqrt{42}}{\sqrt{2}}\right)^{2}+10^{2}\) \(=121 \Rightarrow {f}_{{rms}}=11\, {A}\)
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