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CBSE 12th Standard Physics Subject Alternating Current Ncert Exemplar 1 Mark Questions 2021

Published on: 23/05/2021

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Std: 12th Standard CBSE - Physics

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Questions:

Generic 2

4 x 2 = 8

A train is moving with uniform speed from North and South, why will any induced emf appear across the ends of its axle?

Consider a magnet surrounded by a wire with an ON/OFF switch in the figure. If the switch is thrown from the OFF position (open circuit) to the ON position (closed circuit), will a current flow in the circuit? Explain.

The magnetic flux linked with a uniform surface area A in a uniform magnetic field is given by \(\phi \) = B.A = BA cos \(\theta \) .So flux linked will change, only when either B or A or the angle between B and A change.

The alternating current in a circuit is described by the graph shown in the figure. Find the rms current in this graph.

The current through an AC circuit is \({ I }_{ t }={ I }_{ 0 }\left( t/\tau \right) \) for sometime. Determine the rms current through the circuit over time interval t = 0 to t =\(\tau\).

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Answers Key:

Generic 2

4 x 2 = 8

As the train is intercepting vertical components of the earth's magnetic field.The flux linked with the changes.So, there will be an induced emf.

When the switch is thrown from the OFF position (open circuit) to the ON position (closed circuit), then neither B nor A and the angle between B and A does not change.Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently, no current will flow in the circuit.

From the graph, I₁ = 1 A, I₂ = -2 A and I₃ = 1A
\(\begin{aligned} I_{\mathrm{rms}} & =\sqrt{\frac{I_1^2+I_2^2+I_3^2}{3}}=\sqrt{\frac{1^2+(-2)^2+1^2}{3}} \\ \end{aligned}\)
\(\begin{aligned} =\sqrt{\frac{6}{3}}=\sqrt{2}=1.414 \mathrm{~A} \end{aligned}\)

The mean square current is
\(\begin{aligned} {\overline{I^2}}=\frac{1}{\tau} \int_0^\tau I_0^2(t / \tau)^2 d t \end{aligned}\)
\(\begin{aligned} =\frac{I_0^2}{\tau^3} \int_0^\tau t^2 d t=\frac{I_0^2}{\tau^3}\left[\frac{t^3}{3}\right]_0^\tau=\frac{I_0^2}{3} \quad[\because t=\tau] \\ \end{aligned}\)
\(\begin{aligned} \therefore \quad I_{\text {rms }} & =\sqrt{\overline{I^2}}=\sqrt{\left(\frac{I_0^2}{3}\right)}=\frac{I_0}{\sqrt{3}} \end{aligned}\)