GATE EE 2001

In a series RLC circuit, at resonance
V_L=jQV_{source} and V_c=-jQV_{source}
also for Q \gt 1, |V_c| \gt |V_{source}|
Hence, option (C) is correct.

\because \;\;P\propto \frac{1}{R}
Therefore , resistance of 40 W bulb \gt resistance of 60 W bulb.
For series connection, current through both the bulbs will be same P=I^2R (for series connection).
Power consumed by 40 W bulb \gtPower consumed by 60 W bulb.
Hencem the 40 W bulb glows brighter.

At t=0^+ inductor works as open circuit. Hence, complete source voltage drops across it and consequently, current through the resistor R is zero. Hence, voltage across the resistor at t=0^+ is zero. And further with time it rises accroding to V_R(t)=(1-e^{-Rt/L})u(t).

M=K\sqrt{L_1L_2}
where , K= coefficient of coupling
\because \; 0 \lt K \lt 1
\therefore \; M \leq \sqrt{L_1L_2}

For a passive two port network, output powe can never be grater than input power.

When the value of input voltage is maximum, rate of change of core flux is minimum, as both are 90^{\circ} out of phase in case of sinusoidal input.

For fifth harmonics, 5\beta =180^{\circ}
\Rightarrow \;\;\beta =36^{\circ}