At time , terminal A in the circuit shown in the figure is connected to B by a key and an alternating current with and starts flowing in it with the initial direction shown in the figure. [2014]

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 $I (t) = I_{0} \cos (ω t),$ with $I_{0} = 1 A$ and $ω = 500 {rad s}^{- 1}$ starts flowing in it with the initial direction shown in the figure. At $t = \frac{7 π}{6 ω},$ the 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 $C = 20 μ F,$ $R = 10 Ω$ and the battery is ideal with emf of 50 V, identify the correct statement(s). [2014]

Smart Achievers · Accepted Answer

(3, 4)

For maximum charge on the capacitor, $\frac{d Q}{d t} = I = 0$

$I = I_{0} \cos ω t = \cos 500 t$

Till $t = \frac{7 π}{6 ω}$ , the charge will be maximum at $t = \frac{π}{2 ω}$

$Q' = \int_{0}^{π / (2 ω)} \cos 500 t d t = {[\frac{\sin 500 t}{500}]}_{0}^{π / 2 ω}$

$= \frac{1}{500} \sin (500 \times \frac{π}{2 \times 500}) = \frac{1}{500} C$

i.e., $Q_{\max} = \frac{1}{500} C = 2 \times 10^{- 3} C$

From the graph it is clear that just before $t = \frac{7 π}{6 ω},$ the current is in the anticlockwise direction. Immediately after A is connected to D.

At $t = \frac{7 π}{6 ω},$ the charge on the upper plate of the capacitor is

$\int_{0}^{7 π / (6 ω)} \cos 500 t d t = \frac{1}{500} \sin (500 \times \frac{7 π}{6 \times 500})$

$= - \frac{1}{500} \times \frac{1}{2} = - 10^{- 3} C$

Now applying KVL,

$50 + \frac{10^{- 3}}{20 \times 10^{- 6}} - 10 i = 0$ $\Rightarrow i = 10 A$

The maximum charge on $C$ , $Q = C V = 20 \times 10^{- 6} \times 50 = 10^{- 3} C$

Therefore, the total charge flown from the battery is $2 \times 10^{- 3} C$

Solution

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

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

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

4 $Q = 2 \times 10^{- 3} C$

Want Us to Email you About Special Offers & Updates?

Solution

1 Magnitude of the maximum charge on the capacitor before t=7π6ω is 1×10-3 C

2 The current in the left part of the circuit just before t=7π6ω is clockwise.

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

4 Q=2×10-3 C

Want Us to Email you About Special Offers & Updates?

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

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

4 $Q = 2 \times 10^{- 3} C$