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**Q 33. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combinations is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances? [2018](1) 990 Ω(2) 505 Ω(3) 550 Ω(4) 910 ΩAns:** (3)

(i)

(ii)

From (i) and (ii)

⇒ l = 55 cm

(1)

and R1 + R 2 = 1000 Ω (iii)

from (i) and (iii)

R = 550 Ω

**Q 34. In the given circuit all resistances are of value R ohm each. The equivalent resistance between A and B is: [2018](1) 5R/3(2) 3R(3) 5R/2(4) 2RAns:** D

Due to short circuit current will flow along

B-C-D-E-F-A

So

R eq = R + R

R eq = 2R

(1) 19.15 Ω and 60.5 cm

(2) 19.15 Ω and 39.5 cm

(3) 8.16 Ω and 39.5 cm

(4) 8.16 Ω and 60.5 cm

Ans:

For balanced wheat stone bridge

x(100 − l

So x(100 − 39.5) = 12.5(39.5)

x = 8.16 Ω

If x and y inter changed

y(100 − l

12.5(100 − l

get l

[2018]

(1) 3.3mA

(2) 2.5mA

(3) 5.5mA

(4) 6.7mA

Ans:

(1) 1.25 × 10

(2) 2.5 × 10

(3) 2.5 × 10

(4) 1.25 × 10

Ans:

(1) 550Ω

(2) 220Ω

(3) 55Ω

(4) 110Ω

Ans:

S = 110Ω

[2018]

(1) C

(2) C

(3) C

(4) C

Ans:

q = C

Both capacitor will have same charge as they are connected in series.

(1) A rheostat can be used as a potential divider

(2) Kirchhoff’s second law represents energy conservation

(3) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude

(4) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed [2017]

Ans:

Solution:

In a balanced Wheatstone bridge, the null point remains unchanged even if cell and galvanometer are interchanged.

(1) 2.535 × 10

(2) 4.005 × 10

(3) 1.985 × 10

(4) 2.045 × 10

Ans:

Solution:

ig = 5 × 10

G = 15 Ω

Let series resistance be R.

V = i

10 = 5 × 10

R = 2000 – 15 = 1985 = 1.985 × 10

In the above circuit the current in each resistance is [2017]

(1) 0.5 A

(2) 0 A

(3) 1 A

(4) 0.25 A

Ans:

Solution:

The potential difference in each loop is zero.

∴ No current will flow.

[2017]

(1)

(2)

(3)

(4)

Ans:

When key is at point

V

when key is at (3)

V

** **

**Choose correct statement. [2017](1) Reading of A1 is 18 A(2) Reading of V is 9 V(3) Reading of V is 7 V(4) Reading of A1 is 2 AAns:** (4)

Initially

Finally

R

R

R

P

(1) 240 N/C

(2) 360 N/C

(3) 420 N/C

(4) 480 N/C

Ans:

Charges on 3 μF = 3 μF x δV = 24 μC∴ Charge on 4 μF = Charge on 12 μF = 24 μC

Charge of 3 μF = 3 μF ´ 2V = 6 μC

Charge of 9 μF = 9 μF ´ 2V = 18 mC

Charge on 4 μF + Charge on 9 μF = (24 + 18) μF = 42 μC

∴ Electric field at 30 m = 9 x 10

(1) 0.01 Ω

(2) 2 Ω

(3) 0.1 Ω

(4) 1 Ω

Ans:

Maximum voltage that can be applied across the galvanometer coil = 100 Ω x 10

If R

R

⇒ R

(1) all in series

(2) two in parallel and one in series

(3) two in series and one in parallel

(4) all in parallel

Ans:

(2016)

(1) 1

(2) 3/4

(3) 1/4

(4) 1/2

Ans:

to get maximum heat generation from r.

r = R

i

substituting i

S(R+G) = RG

(1) r

(2) r

(3) r

(4) r

Ans:

Its clear, shunt should be applied in parallel and least the shunt resistance, better the ammeter is.

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