At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I (t) = I0 cos (ωt), with I0 = 1A and ω = 500 rad s–1 starts flowing in it with the initial direction shown in the figure. At , they key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 µF, R = 10 and the battery is ideal with emf of 50V, identify the correct statement(s).

Engineering

Physics

AC Circuit Analysis

Question

At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I (t) = I₀ cos (ωt), with I₀ = 1A and ω = 500 rad s^–1 starts flowing in it with the initial direction shown in the figure. At $t = \frac{7 π}{6 ω}$ , they key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 µF, R = 10 $Ω$ and the battery is ideal with emf of 50V, identify the correct statement(s).

Magnitude of the maximum charge on the capacitor before $t = \frac{7 π}{6 ω}$ is 1 × 10^–3 C.

Q = 2 × 10^–3 C.

Immediately after A is connected to D, the current in R is 10 A.

The current in the left part of the circuit just before $t = \frac{7 π}{6 ω}$ is clockwise.

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$I = I_{0} cos ω t = cos (\frac{7 π}{6 ω}) = - \frac{\sqrt{3}}{2}$

So direction of current is anticlockwise.

$Q = \int Idt = \int_{0}^{\frac{7 π}{6 ω}} cosωt dt = 1 0^{- 3} C$ (Lower plate positive)

Final charge on capacitor = 20 × 10^–6 × 50 = 10^–3 C (Upper plate positive)

So, max. charge flown = $\frac{I_{0}}{ω}$ = 2 × 10^–3 C

$V = \frac{10^{- 3}}{20 \times 10^{- 6}} = \frac{10^{3}}{20} = \frac{100 \times 10}{20} = 50 V$

So, current = $\frac{100}{10}$ = 10 A