#### For driving a current of 3 A for 5 min in an electrical circuit, 900 J of work is to be done. Then the emf of the source in the circuit will be ____.

q = I t =3×5×60 = 900 C

∴V = W/ q = 900/900 = 1 V

#### Wheatstone bridge is better than other methods of measuring resistance because ____.

Usually, null methods are always more accurate than deflection method. Therefore, Wheatstone bridge is better than other methods of measuring resistance.

#### How many electrons pass through a wire in 1 min if the current passing through the wire is 200 mA?

e = 1.6 × 10^{19} C

I = 200 mA = 200 × 10^{-3} A

t = 1 min = 60 s.

n = It/e

= 200×10^{-3}×60/1.6×10^{-19}= 7. 5×10^{19}

#### When an electric charge 100 C is flowing across a conductor in 5 min, the strength of current is

Q = 100 C, t = 5 min = 5 × 60 = 300 s. Therefore, I =Q/t = 100/300=0.33 A.

#### If two similar cells are connected first in series, and then in parallel, then the ratio of balancing length in the potentiometer wire will be:

When cells are connected in series, potential emf = 2E

When cells are connected in parallel, emf = E

∴l_{2}/l_{1} = 2E/E = 2: 1

#### If a current of 4.8 A is flowing in a copper wire of cross-sectional area 3 x 10-4m2, then the current density of the wire is ____.

I = 4.8; A = 3 x 10^{-4}m^{2}

So, current density (j) = 4.8/3 x10^{-4} =1.6×10^{4}Am^{-2}

#### The sensitivity of potentiometer can be increased by ____.

For a given potential difference, the sensitivity of potentiometer increases with the increase in the length of the potentiometer wire. Greater the length of the wire, smaller will be the potential gradient, and hence greater will be the sensitivity of the potentiometer.

#### 2 cells have emf of 2.0 V and 4.0 V and internal resistance of 0.10 Ω and 0.20 Ω respectively.

The equivalent emf of the 2 cells when they are connected in opposing series is (4 - 2) = 2 V.

The equivalent internal resistance of the 2 cells is 0.30 Ω. This is in series with the external resistance. Hence, resistance in the circuit is 2.7 + 0.3 = 3 Ω.

Therefore, I = 4 – 2/3 = 2/3 A.

#### If a battery of emf 6 V and internal resistance 1 Ω is connected to a resistor and the current is 0.5 A, then the resistance of the resistor will be ____.

ε = 6 V, r = 1 Ω, I = 0.5 A

I = ε/R + r

R = (ε/I) - r =( 6/0.5) -1 =11 Ώ

#### If 6.25 × 10^{18} electrons flow through a given cross section in unit time, then the current is ____. (charge of an e- is 1.6 × 10^{-19} C)

n = 6.25 × 10^{18}, e = 1.6 × 10 ^{-19} C, t = 1s.

Therefore, I = Q/t = ne/t

= 6. 25×10^{18}×1.6×10^{-19 }/1=10×10^{-1} = 1 A

#### In a Wheatstone bridge, if the galvanometer shows zero deflection, the unknown resistance will be____ Ω. (Given P = 1000 Ω, Q = 10000 Ω and R = 20 Ω)

According to Wheatstone bridge,

P/Q = R/S

⇒1000/10000 = 20/S

Therefore, S = 200 Ω

#### If a cell has an emf 5 V, and an internal resistance 0.10 Ω is connected to a resistance of R Ω, the current from the cell is 2A, then the voltage drop across R will be ____.

I = εR+r = 59+1 = 1/2 A

V = ε - I r = 5 – 1/2×1=4.5 V

#### If a potential difference of 6 V is applied across a conductor of length 0.12 m, electron mobility is 5.6 x 10^{-6} m^{2} V^{-1} s^{-1}, then the drift velocity of electrons will be ____.

E = V/l = 6/0.12 = 50 Vm^{-1}

Now, V_{d }= µ E = 5.6 x 10^{-6}x 50

= 2.8 x 10^{-4} m s ^{-1}

#### If a cell of emf 5 V and of negligible internal resistance is connected in series with a potentiometer wire of length 500 cm. the emf of Leclanche cell is found to balance on 250 cm of the potentiometer wire, then the emf of Leclanche cell will be _____.

The fall of potential per unit length of potentiometer wire =5/500

= 10^{-2}vcm^{-1}

As Leclanche cell is balanced against 250 cm

emf of Leclanche cell = 250 x 10^{-2} = 2.5 V

#### Kirchoff's current law obeys conservation of ____.

Kirchoff's current law obeys law of conservation of charge.

#### In potentiometer, a balance point is obtained when ____.

In potentiometer when the potential difference of the wire between the positive end of the battery to jockey becomes equal to the emf of the experimental cell a balance point is obtained.

#### If each cell has an emf of 2.0 V and negligible internal resistance, the current maintained in a 3 Ω resistor by arranging 3 cells in series and 3 cells in parallel is ____.

When cells are in series, the equivalent emf of the 3 cells = 3 × 2 V; ∴I =6/3 =2A

When the cells are in parallel, the equivalent emf of the 3 cells = 2V; ∴I = 2/3 = 0.67A

#### The flow of electric charge through the conductor is called ____.

The flow of electric charge through the conductor is called current.

#### If 6 V battery is used to supply a resistance R_{1} in parallel with R_{2} and the current through R_{1} is 6 A and that through R_{2} is 1.5 A, then the load on the battery is ____.

I = (6 + 1.5) = 7.5 A ∴ P = V I = 6 x 7.5 = 45 W

#### In a potentiometer experiment, the balancing with a cell is at length 120 cm. On shunting the cell with a resistance 3 Ω, the balancing length becomes 60 cm. Then the internal resistance of the cell would be

The internal resistance of the cell,

r = ( l_{1 }/ l_{2} )- 1 s

l_{1} = 120 cm, l_{2}= 60 cm and s = 3 Ω

r = (120/60) - 1×3

= 3 Ω

#### If a series combination of 3 resistors takes a current of 2A from a 24 V supply and if the resistors are in the ratio of 1: 2:3, then the values of unknown resistors respectively are ____.

E = 24 V, I = 2A

Rs =24/2 = 12 Ώ

Let R_{1}, R_{2} and R_{3} be 3 unknown resistors.

Let
R_{1} = R , R_{2}= 2R and R_{3} =3R

R_{s}=R_{1}+ R_{2}+R_{3}

12 = R + 2R + 3R

6 R = 12

R = 2

∵R_{1}:R_{2}:R_{3} = 1:2:3 ; and R = 2 i.e. R_{1} = 2

R_{1}:R_{2}:R_{3} = 2:4:6 i.e. 2Ω: 4Ω : 6Ω

#### If 60 W, -120 V bulb is to be operated on 80 V, then the current in the bulb is ____.

V = 120 V, P = 60 W

R = V^{2}/P = 120×120/60 = 240 Ώ

So, I = V/R = 80/240 = 1/3 A= 0.33 A

#### If a 100 W, 200 V bulb is connected to a 160 V supply, then the power consumption would be ____.

V = 200 V, P =100 W

Resistance R =V^{2}/P = 200×200/100 = 400 Ώ

So, P = V^{2}/R = 160×160/400 (as V=160)= 64 W

#### With a certain cell, the balance point is obtained at 65 cm from the end of a potentiometer wire. With another cell whose emf differs from that of the first by 0.1 V, the balance point is obtained at 60cm. Then the emf of each cell is ____.

Let E_{1} and E_{2} = emf of the two cells respectively. l_{1} = 65 cm and l_{2} = 60 cm

As balancing length for the 2nd cell is less than that for the 1st cell, the emf of the 2nd cell is less than that of the 1st cell, i.e. E_{2 }< E_{1}.

As the emf of 2 cells differ by 0.1 V,

E_{2} = E_{1} - 0.1

Now, E_{1}/E_{2} = l_{1}/l_{2}

E_{1}/E_{1}=65/60

∴ E_{1}= 1.3 V

E_{2} = 1.3 - 0.1 = 1.2 V

#### If a battery consisting of 2 cells of 1.45 V and internal resistance of 0.15 Ω each is sending a current through the filament of the lamp having resistance of 2.5 Ω, then the current is ____.

I = nE/R+nr, where n = 2, E = 1.45 V, R = 2.5 Ω and r = 0.15 Ω.

∴I = 2×1.45/2.5 + 2×0.15 = 2.9/2.8 = 1.03 A

#### If a generator is supplying power to a factory by cables of resistance 20Ω and the generator is generating 50 kW power at 5000 V, then the power received by the factory would be ____.

Current I = 50×1000/5000 = 10 A

#### If in the experiment of Wheatstone bridge, the battery and galvanometer are interchanged, then balance points will ____.

The balance points remain unchanged when the battery and galvanometer of a Wheatstone bridge are interchanged. Sensitivity of the bridge changes.

#### If a parallel combination of 3 resistors takes a current of 5 A from a 20 V supply, and if the two resistors are 10 Ω and 8 Ω, then the value of the 3rd will be ____.

E = 20 V I = 5A

∴Rp = 20/5 = 4 Ώ

Let R = unknown resistance.

1/Rp = 1/R + 1/10 + 1/8

i.e.1/4 = 1/R + 1/10 + 1/8

1/R = 1/4 – 1/10 – 1/8 = 1/40

Therefore, R = 40 Ω

#### If a charge of 60 C passes through an electric lamp in 2 min, then the current in the lamp is

Q = 60 C, t = 2 min = 2 × 60 = 120 seconds

I = Q/t

Therefore, I = 60/120 = 1/2 = 0.5 A.

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