A2 June 2023 Paper 1 Q9
9 In an electrical circuit, the alternating current \(I\) amps is given by \(I = a\sin nt\), where \(t\) is the time in seconds and \(a\) and \(n\) are positive constants. The RMS value of the current, in amps, is defined to be the square root of the mean value of \(I^2\) over one complete period of \(\dfrac{2\pi}{n}\) seconds.
Show that the RMS value of the current is \(\dfrac{a}{\sqrt{2}}\) amps. [6]
| Scheme | Marks | AO |
|---|---|---|
| \(k\displaystyle\int_0^{\frac{2\pi}{n}}\sin^2 nt\,\mathrm{d}t = \frac{1}{2}k\int_0^{\frac{2\pi}{n}}(1 - \cos 2nt)\,\mathrm{d}t\) | M1* A1 | 3.1a 1.1 |
| \(= \dfrac{1}{2}k\left[t - \dfrac{1}{2n}\sin 2nt\right]_0^{\frac{2\pi}{n}}\) | M1dep | 1.1 |
| \(= \dfrac{1}{2}a^2\dfrac{2\pi}{n} = \dfrac{\pi a^2}{n}\) | A1 | 1.1 |
| mean value of \(I^2 = \dfrac{\pi a^2}{n} \Big/ \dfrac{2\pi}{n} = \dfrac{a^2}{2}\) | M1 | 1.1 |
| \(\Rightarrow \text{RMS value} = \frac{a}{\sqrt{2}}\) | A1 | 2.2a |
| [6] |
Notes
M1*: use of double angle formula for \(\sin^2 nt\)
M1dep: \(\left[t - \frac{1}{2n}\sin 2nt\right]\)
A1: (2nd) www
M1: (2nd) for dividing by \(\frac{2\pi}{n}\)
A1: (3rd) www AG
Alternative method
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{\left(\frac{2\pi}{n}\right)}k\displaystyle\int_0^{\frac{2\pi}{n}}\sin^2 nt\,\mathrm{d}t = \frac{n}{2\pi}k\int_0^{\frac{2\pi}{n}}\sin^2 nt\,\mathrm{d}t\) | M1 |
| \(= \dfrac{1}{2} \times \dfrac{n}{2\pi}k\displaystyle\int_0^{\frac{2\pi}{n}}(1 - \cos 2nt)\,\mathrm{d}t\) | M1* A1 |
| \(= \dfrac{n}{4\pi}k\left[t - \dfrac{1}{2n}\sin 2nt\right]_0^{\frac{2\pi}{n}}\) | M1dep |
| \(= \dfrac{a^2n}{4\pi}\left(\dfrac{2\pi}{n}\right) = \dfrac{a^2}{2}\) | A1 |
| \(\Rightarrow \text{RMS value} = \frac{a}{\sqrt{2}}\) | A1 |
M1: for dividing their integral by \(\frac{2\pi}{n}\)
M1*: use of double angle formula for \(\sin^2 nt\)
M1dep: \(\left[t - \frac{1}{2n}\sin 2nt\right]\)
A1: (2nd) www
A1: (3rd) www AG