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1

### JEE Main 2021 (Online) 1st September Evening Shift

For the given circuit the current i through the battery when the key in closed and the steady state has been reached is __________.

A
6 A
B
25 A
C
10 A
D
0 A

## Explanation

We know in study state potential difference across inductor = 0

So, equivalent circuit becomes

$${1 \over {{R_{eq}}}} = {1 \over 3} + {1 \over 3} + {1 \over 3} = 1$$

$$\Rightarrow$$ Req = 1$$\Omega$$

$$\Rightarrow$$ Circuit becomes

$$\Rightarrow i = {{30} \over 3} = 10$$ A
2

### JEE Main 2021 (Online) 31st August Evening Shift

Statement - I :

To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load RL.

Statement - II :

To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RL.

In the light of the above statements, choose the most appropriate answer from the options given below :
A
Statement I is true but Statement II is false
B
Statement I is false but Statement II is true
C
Both Statement I and Statement II are false
D
Both Statement I and Statement II are true

## Explanation

To convert pulsating dc into steady dc both of mentioned method are correct.
3

### JEE Main 2021 (Online) 31st August Morning Shift

In an ac circuit, an inductor, a capacitor and a resistor are connected in series with XL = R = XC. Impedance of this circuit is :
A
2R2
B
Zero
C
R
D
R$$\sqrt 2$$

## Explanation

$$Z = \sqrt {{{({X_L} - {X_C})}^2} + {R^2}} = R$$ $$\because$$ XL = XC

Option (c)
4

### JEE Main 2021 (Online) 31st August Morning Shift

A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :
A
$${{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}$$
B
$${{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over a}$$
C
$${{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{b^2}} \over a}$$
D
$${{{\mu _0}} \over {4\pi }}{{8\sqrt 2 } \over b}$$

## Explanation

$$B = \left[ {{{{\mu _0}} \over {4\pi }}{I \over {b/2}} \times 2\sin 45} \right] \times 4$$

$$\phi = 2\sqrt 2 {{{\mu _0}} \over \pi }{I \over b} \times {a^2}$$

$$\therefore$$ $$M = {\phi \over I} = {{2\sqrt 2 {\mu _0}{a^2}} \over {\pi b}} = {{{\mu _0}} \over {4\pi }}8\sqrt 2 {{{a^2}} \over b}$$

Option (a)

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