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**Q 1. The current I _{1} (in A) flowing through 1 Î© resistor in the following circuit is [2020](1) 0.4 (2) 0.5 (3) 0.2 (4) 0.25 **

We have the following circuit figure

Two 1 Î© resistors are connected is parallel so equivalent resistance is

So, circuit becomes

Equivalent resistance of 0.5 Î© and 2 Î© resistor connected in series is

R' = 0.5 + 2 - 2.5 Î©

So, circuit becomes

Equivalent resistance of circuit is

So current in the circuit is

I = V/R = 9/10A

Then

Potential difference across 0.5 Î© resistor is

V' = i

So, current through 1 Î© resistor is

**Q 2. In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric mains is 220 V. The minimum fuse capacity (rated value) of the building will be [2020](1) 10 A (2) 25 A (3) 15 A (4) 20 A Ans: **(4)

Given that, 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W, 2 heaters of 1 kW = 1000 W, Voltage of electric main, V = 220 V

Total power is given by

P = (15 Ã— 45) + (15 Ã— 100) + (15 Ã— 10) + (2 Ã— 1000)

= 675 + 1500 + 150 + 2000 = 4325 W

We know that,

I = P/V = 4325/220 = 19.659 âˆ¼ 20A

so minimum fuse capacity of the building will be 20 A.**Q 3. The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 Î© is connected in parallel to the cell, the balancing length changes by 60 cm. If the internal resistance of the cell is ( N/10) Î©, where N is an integer then value of N is ________. [2020]Ans:** 12

balancing length, L = 560 cm

external resistance, R = 10 Î©

New balancing length after external resistance is connected, L' = 60 cm

Internal resistance of the cell, r = N/10Î©

Let emf of cell be E, and potential difference be V

Then, E = 560 V ...(1)

when external resistance connected in parallel, new emf is

From Eq. (1), we have

â‡’ 56= 5(10 + r)

â‡’ 56 = 50+5r

â‡’ 5r = 6 â‡’ r = 1.2 Î©

we have,

r = N/10

â‡’ 1.2 = N/10

â‡’ N = 12

So value of integer N is 12.

Given that L = 1200 cm, I = 60mA, Îµ = 5V, r = 20Î©, l = 1000 cm

We know that

So,

Given that P' = 15Î©, Q' = 12Î©, R' = 4Î©, S' = 10Î©

Condition for balanced Wheatstone bridge is

â‡’ PS = QR

So,

Given condition can be shown by the following circuit figure

Given that

E

We have

V

â‡’ 0 = 10 - 20l

â‡’ I = 0.5A

V

= 10 - (0.5 x 5) = 7.5 a

I = I

V = V

From Eqs. (1) and (2), we have

[2020]

(1) 0.4 A

(2) 2 A

(3) 4 A

(4) zero

Ans:

Total current in the circuit is given by

Effective resistance of the combination is

Given that V = 20 V. So,

I = 20/2 = 10 A

Applying KVL in loop 1, we get

I x I

â‡’ 5I

Applying KVL in loop 2, we get

[ 2 x (l

â‡’ 5l

Ans:

When meter bridge is balanced, we have

Now,

So, new resistance is

So, new balanced point is

**(1) 270 Î©, 10%(2) 27 kÎ©, 10%(3) 27 kÎ©, 20%(4) 270 Î©, 5%Ans: **(2)

This resistor, read from left to right, has the colored bands of RED, VIOLET, ORANGE and SILVER.

The resistance is,

R = (1

R = (RED Ã— 10 + VIOLET) Ã— ORANGE = (2 Ã— 10 + 7) Ã— 1,000

= 27,000 Î© = 27 kÎ©

Since, the last band is silver, the tolerance is 10%.

**Q 10. Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm ^{2}, is v. If the electron density in copper is 9 Ã— 10^{28}/m^{3} the value of v in m/s is close to (take charge of electron to be 1.6 Ã— 10^{âˆ’19} C) [2019]** (1)

(1) 0.02

(2) 3

(3) 2

(4) 0.2

Ans:

Current flows in a copper wire, I = 1.5 A

Charge, q = 1.6 Ã— 10

(1) 2.2%

(2) 2.5%

(3) 1.0%

(4) 0.5%

Ans:

We know that,

Al = Volume (constant)

Now, percentage change is

(1) 3 A

(2) 5 A

(3) 4 A

(4) 2 A

Ans:

Let V be the voltage at C.

According to the Kirchhoffâ€™s current law, we have

** **

**(1) 530 kÎ© Â± 5%(2) 5.3 MÎ© Â± 5%(3) 6.4 MÎ© Â± 5%(4) 64 kÎ© Â± 10%Ans:** (1)

Resistor code for Orange = 3

Resistor code for Yellow = 10

Tolerance for Golden = Â±5%

Therefore, G O Y Golden

**(1) 300 Î©(2) 450 Î©(3) 550 Î©(4) 230 Î©Ans:** (1)

The voltage across resistance R

V_{4} = 5V

Thus, current i_{1} = 5/500 = 0.01 A

Voltage across resistance R_{3} is V_{3} = 0.01 Ã— 100 = 1 V

Thus, Total voltage = V_{1} + V_{3} + V_{4}

(since voltage across R_{2} is same as voltage across R_{3} and R_{4})

â‡’ 18 = 1 + 5 + V_{1}

â‡’ V_{1} = 12 V

â‡’ i = 12/400 = 0.03 A

â‡’ i_{2} = i â€“ i_{1} = 0.03 â€“ 0.01 = 0.02 A

therefore,**Q 15. A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf Îµ and internal resistance r. A cell C having emf Îµ/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure, for which galvanometer shows no deflection is [2019]Ans:** (3)

Solution:

Let at x distance it will show no deflection then current

In case of no deflection

(1) 4 Î©

Ans:

It is bent into an equilateral triangle and the length of each side is same.

And its resistance = 6 Î©

Now, R

**(1) 1, 2(2) 2, 2(3) 0.5, 0(4) 0, 1Ans:** (3)

Potential difference across R

Current i through

Potential difference across R

Therefore, current i through R_{2} = 0**Q 18. A 2W carbon resistor is color coded with green, black, red and brown, respectively. The maximum current which can be passed through this resistor is [2019](1) 20 mA(2) 100 mA(3) 0.4 mA(4) 63 mAAns:** (1)

Resistor from color coding

R = 50 Ã— 10^{2} Î© = 5000 Î©

Power = 2 W Power is P = i^{2}R

â‡’ 2 = i^{2} Ã— 5000

**Q 19. The Wheatstone bridge shown in figure here, gets balanced when the carbon resistor used as R _{1} has the color code (orange, red, brown). The resistors R_{2} and R_{4} are 80 Î© and 40 Î©, respectively. Assuming that the color code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R_{3}, would be [2019]** (1)

(1) Brow n, Blue, Brown

(2) Brown, Blue, Black

(3) Red, Green, Brown

(4) Grey, Black, Brown

Ans:

Color code of Red = 2

Color code of Brown = 10

So, R

The balanced condition of wheat stone bridge is

Hence, the color code is Brown, Blue, Brown.

**(1) 600 Î©(2) 570 Î©(3) 35 Î©(4) 350 Î©Ans:** (2)

(1) 11 Ã— 10

(2) 11 Ã— 10

(3) 11 Ã— 10

(4) 11 Ã— 10

Ans:

**(1) 401.5 Î©(2) 404.5 Î©(3) 403.5 Î©(4) 402.5 Î©Ans:** (4)

Balanced condition of Wheatstone bridge is,

Now

(1)

Ans,

(2)

From Eq. (1) and Eq. (2), we get

(1) 30 W

Ans:

Solution:

In parallel condition, power consumed is

(1) 0.3 V

(2) 0.5 V

(3) 0.6 V

(4) 0.4 V

Ans:

Let null point J be at l cm from A.

Then at balance condition we have

**(1) 20 Î©(2) 40 Î©(3) 60 Î©(4) 30 Î©Ans:** (3)

Consider the following circuit,

Now using Wheatstone bridge equation, we have

Thus, R

Now,

**(1) 1 V(2) 2 V(3) 3 V(4) 6 VAns:**(2)

**Q 27. An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Î©. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is [2019](1) 490 Î©(2) 480 Î©(3) 395 Î©(4) 495 Î©Ans:** (3)

Let i be the current flowing in the wire

If resistance of 10 m length of wire is r, then

Potential difference, Î”V = ir

(1) 0.2 m

(2) 0.3 m

(3) 0.25 m

(4) 0.35 m

Ans:

For zero deflection,

Substitute the value of R

(1) P

Ans:

Solution:

Suppose R

Current

Ans:

Change on capacitor C before turning the switch S

Q = CE (1)

Since, when switch S is turned from A to B charge will be transferred from one capacitor to another capacitor until potential will not same.

Let common potential is Îµ'

Energy dissipated, E = E_{i} âˆ’ E_{f}**Q 30. In the given circuit diagram, the currents, I _{1}= âˆ’0.3 A, I_{4} = 0.8 A and I_{5} = 0.4 A, are flowing as shown. The currents I_{2}, I_{3} and I_{6}, respectively, are [2019]**

**(1) 1.1 A, âˆ’0.4 A, 0.4 A(2) 1.1 A, 0.4 A, 0.4 A(3) 0.4 A, 1.1 A, 0.4 A(4) âˆ’0.4 A, 0.4 A, 1.1 AAns:** (2)

From using KCL at point S

(1) 11.6 V and 11.7 V

(2) 11.5 V and 11.6 V

(3) 11.4 V and 11.5 V

(4) 11.7 V and 11.8 V

Ans:

The circuit may be drawn as shown in the figure.

(1) 1Î©

(2) 1.5Î©

(3) 2Î©

(4) 2.5Î©

Ans:

When the cell is shunted by a resistance of 5 Î©

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