The alternating current in a circuit is described by the graph shown in figure. Show rms current in this graph.
The alternating current in a circuit is described by the graph shown in figure. Show rms current in this graph.
From graph diagram maximum current $\mathrm{I}_{1}=1 \mathrm{~A}$ and $\mathrm{I}_{2}=-2 \mathrm{~A}$ $\therefore$ Average maximum current $\mathrm{I}_{\mathrm{m}}=\sqrt{(1)^{2}+(-2)^{2}}$
$\therefore \mathrm{I}_{\mathrm{m}}=\sqrt{5} \mathrm{~A}$
Now, $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}$
$=\frac{\sqrt{5}}{\sqrt{2}}=\sqrt{\frac{5}{2}}=\sqrt{2.5}=1.58 \mathrm{~A}$
$\therefore \mathrm{I}_{\mathrm{rms}} \approx 1.6 \mathrm{~A}$ is shown in below graph.
Similar Questions
The instantaneous voltages at three terminals marked $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are given by
$V_x=V_0 \sin \omega t$ $V_y=V_0 \sin \left(\omega t+\frac{2 \pi}{3}\right) \text { and }$ $V_z=V_0 \sin \left(\omega t+\frac{4 \pi}{3}\right)$
An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points $\mathrm{X}$ and $\mathrm{Y}$ and then between $\mathrm{Y}$ and $\mathrm{Z}$. The reading(s) of the voltmeter will be
$[A]$ $\mathrm{V}_{\mathrm{XY}}^{\mathrm{mms}}=\mathrm{V}_0 \sqrt{\frac{3}{2}}$
$[B]$ $\mathrm{V}_{\mathrm{YZ}}^{\mathrm{mms}}=\mathrm{V}_0 \sqrt{\frac{1}{2}}$
$[C]$ $\mathrm{V}_{\mathrm{XY}}^{\mathrm{mms}}=\mathrm{V}_0$
$[D]$ independent of the choice of the two terminals
The instantaneous voltages at three terminals marked $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are given by
$V_x=V_0 \sin \omega t$ $V_y=V_0 \sin \left(\omega t+\frac{2 \pi}{3}\right) \text { and }$ $V_z=V_0 \sin \left(\omega t+\frac{4 \pi}{3}\right)$
An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points $\mathrm{X}$ and $\mathrm{Y}$ and then between $\mathrm{Y}$ and $\mathrm{Z}$. The reading(s) of the voltmeter will be
$[A]$ $\mathrm{V}_{\mathrm{XY}}^{\mathrm{mms}}=\mathrm{V}_0 \sqrt{\frac{3}{2}}$
$[B]$ $\mathrm{V}_{\mathrm{YZ}}^{\mathrm{mms}}=\mathrm{V}_0 \sqrt{\frac{1}{2}}$
$[C]$ $\mathrm{V}_{\mathrm{XY}}^{\mathrm{mms}}=\mathrm{V}_0$
$[D]$ independent of the choice of the two terminals
- [IIT 2017]
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Statement $-2$ : Capacitor blocks $d.c.$ and allows $a.c.$ only.
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