Physics Trick Of Current Electricity For NDA

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Trick Of Current Electricity For NDA

Method of finding Electric current and Electric charge

Rate of flow of charge is called the electric current (i) i.e.,

`i = q/t`

Q 2303167948

A current of `0.5` A is drawn by a filament of an electric bulb for `20` min. The amount electric charge that flows through the circuit is
NDA Paper 2 2013

(A)

` 1C`

(B)

`10 C`

(C)

`600 C`

(D)

`300 C`

Solution:

It is given that

Current `(I) = 0.5 A`

Time `(t) = 20 min = 1200 s`

We know that,

Charge `(Q) = I xx t`

`=0.5 xx 1200=600 C`

Correct Answer is `=>` (C) `600 C`

Q 2238612502

A `60` watt bulb carries a current `0.5` amp. The total charge passing through it in `1` hour is :
BITSAT Mock

(A)

`1800` coulomb

(B)

`2400` coulomb

(C)

`3000` coulomb

(D)

`3600` coulomb

Solution:

The total charge passing through bulb is

`q = it` ...(i)

Here, `i = 0.5 A, t = 1` hour

`t = 1 xx 60 xx 60 sec`

`t = 3600 ` sec

from eq. (i)

`q = 0.5 xx 3600 = 1800` coulomb

Correct Answer is `=>` (A) `1800` coulomb

Method of finding Drift velocity

The current related with drift velocity is `i = n eAv_d`

We can also write `v_d = (i)/(n e A) = (J)/(n e)= (sigmaE)/(n e) = (E)/(rhon e) = V/(rhol n e)`

Q 2448680503

For ohmic conductor the drift velocity `v_d` and the electric field applied across it are related as
BCECE Stage 1 2013

(A)

` v_d prop sqrtE`

(B)

`v_d prop E^2`

(C)

`v_d prop E`

(D)

`v_d prop 1/E`

Solution:

We have `I= n e Av_d`

and `I = V/R`

`:. V/R = n e Av_d`

or ` V = n e Av_d . R`

` = n eAv_d (rho l/A)`

`:. V/l = E = n e rhov_d`

`=> v_dpropE`

Correct Answer is `=>` (C) `v_d prop E`

Ohm's Law

`I propto V` or `V= IR`

where, R is a constant known as the electrical resistance of given conductor.

Q 2313778640

Ohm's law can also be taken as a statement for
NDA Paper 2 2013

(A)

conservation of energy

(B)

conservation of electric charge

(C)

conservation of angular momentum

(D)

non-conservation of momentum of the flowing charges

Solution:

Ohm's law follow the law of conservation of energy.
According to ohm's law,

`text(Electric current) (I) = (text(Voltage) (V))/(text(Resistance) (R))`

Correct Answer is `=>` (A) conservation of energy

Q 2374223156

Ohm's law defines
NDA Paper 2 2013

(A)

a resistance

(B)

current only

(C)

voltage only

(D)

both current and voltage

Solution:

Ohm's law states that the current through a conductor
between two points is directly proportional to the potential
difference across the two points. Introducing the constant of
proportionality, the resistance

`I=V/R` or `V/I=R` (constant)

where, `I->` in ampere, `V ->` in volt, `R ->` ohm (`Omega`)

Correct Answer is `=>` (A) a resistance

Q 2734001852

Which one of the following physical quantities does NOT affect the resistance of a cylindrical resistor?
NDA Paper 2 2017

(A)

The current through it

(B)

Its length

(C)

The resistivity of the material used in the reststor

(D)

The area of cross-section of the cylinder

Correct Answer is `=>` (A) The current through it

Method of finding Specific Resistance

Ratio of electric field and the current density at a point in the conductor is called the specific resistance or resistivity. It is a constant for the material.

`rho= E/J` or `rho = (RA)/l`
Its unit is ohm-metre.

Q 2364391255

Two copper wires A and B of length 1 and 2 respectively, have the same area of cross-section. The ratio of the resistivities of wires A to the B is
NDA Paper 2 2011

(A)

(B)

(C)

(D)

`1/2`

Solution:

Resistivity,` rho=R A/l=> rho prop A/l`

`(rho_A)/(rho_B)=A_A/A_B xx l_B/l_A= 1/1 xx (2l)/l=2`

Correct Answer is `=>` (B) 2

Q 2384580457

The resistance of a wire is `10 Omega`. If it is stretched ten times, the resistance will read
NDA Paper 2 2012

(A)

`1 Omega`

(B)

`10 Omega`

(C)

`100 Omega`

(D)

`1000 Omega`

Solution:

Resistance `= 10^3 = 1000 Omega R prop e^2`

Correct Answer is `=>` (D) `1000 Omega`

Q 2365678565

The specific resistance of a conducting wire depends upon
NDA Paper 2 2010

(A)

length of the wire, area of cross-section of the wire and material o1 the wire

(B)

length of the wire and area of cross-section of the wire but not on the material of the wire

(C)

material of the wire only but neither on the length of the wire nor on the area of cross-section of the wire

(D)

length of the wire only but neither on tile area of cross-section of the wire nor on the material of the wire

Solution:

Specific resistance or resistivity is a property of material and does not depend upon its shape and size, its length, thickness and area of cross-section.

Correct Answer is `=>` (C) material of the wire only but neither on the length of the wire nor on the area of cross-section of the wire

Q 2356291174

The dimensions of a rectangular block of carbon are `2 cm xx 2 cm xx 10 cm`. If the resistivity of the carbon is `2 xx 10^( -5) Omega - m`, what is the resistance measured between the two square surfaces?
NDA Paper 2 2007

(A)

`5 xx 10^(-3) Omega`

(B)

`2 xx 10^(-3) Omega`

(C)

`5 xx 10^(-2) Omega`

(D)

`2 xx 10^(-2) Omega`

Solution:

Here, resistivity, `rho = 2 xx 10^(-5) Omega m`

Breadth of block `= 2 cm = 2 xx 10^(-2) m`

Height of block`= 10 cm= 10 xx 10^(-2) m`

Resistance between the two square surface

= Resistivity `xx (text(Length of rectangle))/(text(Area of cross section of rectangle))`

`= 2 xx 10^(-5) xx ((10 xx 10^(-2)))/((2 xx 10^(-2) xx 2 xx 10^(-2))` `Omega`

`= 5 xx 10^(-3) Omega`

Correct Answer is `=>` (A) `5 xx 10^(-3) Omega`

Q 2386878777

A wire has a resistance of `32 Omega`. It is melted and drawn into a wire of half of its original length. What is the resistance of the new wire?
NDA Paper 2 2008

(A)

`32 Omega`

(B)

`16 Omega`

(C)

`8 Omega`

(D)

`4 Omega`

Solution:

Resistance of wire, `R = 32 Omega`

Resistance, `R = rho l/A`

i.e., `R prop l/A`

or `R prop l^2/(Al)`

or `R=l^2/V`

`:. (R')/R=((l')/l)^2` (given, `V` is the same for both conditions)

`=> (R')/32=((l//2)/l)^2=1/4`

or `R'=32/4=8 Omega`

Correct Answer is `=>` (C) `8 Omega`

Method of finding equivalent resistance

`(i) text(Series combination)` In this combination, resistances are joined end to end.

In series combination, equivalent resistance is equal to sum of individual resistance.
`R =R_1 + R_2 + R_3`

`(ii) text(Parallel combination)` In this combination, first ends of all the resistances is connected to one point and second ends of all the resistances is connected to other point.
In parallel combination, equivalent resistance is given by`1/R = (1)/(R_1) + (1)/(R_2) + (1)/(R_3)`

Q 2303180048

What should be the reading of the voltmeter `V` in the circuit given above?

(All the resistances are equal to `1 Omega` and the battery is of `1.5 V`)
NDA Paper 2 2013

(A)

`1.5 V`

(B)

`0.66 V`

(C)

`1 V`

(D)

`2 V`

Solution:

The given circuit is

Equivalent resistance of the circuit

`=1+( 1 xx 1)/(1+1)`

`=1+1/2=3/2 Omega`

Total voltage `(V) = 1.5 V`

Current ` (I)=V/(R_(eq))`

`=(1.5)/(3//2) =1 A`

Reading of voltmeter `(V) = I xx 1`

`=1 xx 1=1 V`

Correct Answer is `=>` (C) `1 V`

Q 2316891779

If an input current 3 A flows through the circuit shown above, what is the value of the current flowing through the `4 Omega` resistor'?
NDA Paper 2 2007

(A)

1.6 A

(B)

0.8 A

(C)

0.75 A

(D)

0.4 A

Solution:

Equivalent resistance of circuit

`1/R=1/2+1/8+1/4+1/1`

or `R=8/15 Omega`

Now, potential difference of each branch

`V=iR=8/15 xx 3=8/5 V`

Hence, current through ` 4 Omega` resistance

`i=V/R=(8//5)/4=8/20=0.4 A`

Correct Answer is `=>` (D) 0.4 A

Q 2384756657

Consider the following circuit. The current flowing through each of the resistors connected in the above circuit is
NDA Paper 2 2012

(A)

2 A

(B)

1 A

(C)

9 A

(D)

4 A

Solution:

Equivalent resistance of circuit `1/R_(eq)=1/R_1 +1/R_2`

`1/R_(eq) = 1/3 +1/3=2/3`

`R_(eq) = 3/2`

Now `V=IR`

`I=6/(3/2)`

`I=4A`

Since both resistance have same values so current will be equally divided in both resistances

Current in each resistance `I=V/R=6/3=2A`

Correct Answer is `=>` (A) 2 A

Q 2315067869

The effective resistance of three equal resistances, each of resistance `r`, connected in parallel, is
NDA Paper 2 2010

(A)

`3/r`

(B)

`r/3`

(C)

`3r`

(D)

`r^3`

Solution:

In parallel grouping of resistar,ces, the potential difference across each resistance is same but current in the circuit is distributed amongst various resistances in the inverse ratio of their resistance. i.e., `I_1:I_2:I_3=1/R_1 :1/R_2 :1/R_3`

The equivalent resistance for a parallel combination is given as

`1/R=1/R_1+1/R_2-1/R_3`

According to question, `R_1 = R_2 = R_3 = r`

`:. 1/R=1/r+1/r+1/r`

`=> R=r/3`

Thus, in parallel combination, the equivalent resistance is lesser than even the smallest individual resistance.

Correct Answer is `=>` (B) `r/3`

Q 2376712676

Which one of the following is the resistance that must be placed parallel with `12 Omega` resistance to obtain a combined resistance of `4 Omega` ?
NDA Paper 2 2009

(A)

`2 Omega`

(B)

`4 Omega`

(C)

`6 Omega`

(D)

`8 Omega`

Solution:

Let resistance `x` is in parallel with `12 Omega` resistance

Then, `R=(x xx 12)/(x+12)`

`=> 4=(12x)/(x+12)`

`=> 4x+48=12x`

or, `8x=48`

or `x=6 Omega`

Correct Answer is `=>` (C) `6 Omega`

Q 2345712663

Three resistance coils of restiveness `1 Omega, 2 Omega` and `3 Omega` are connected in series. If the combination is connected to a battery of `9 V`, what is the potential drop across the resistance coil of `3 Omega`?
NDA Paper 2 2011

(A)

`2.0 V`

(B)

`3.0 V`

(C)

`4.5 V`

(D)

`6.0 V`

Solution:

Effective resistance, `R = R_1 + R_2 + R_3`

`=1+2+3=6Omega`

According to Ohm's law, `I=V/R=9/6= 1.5A`

`:.` Potential drop across the resistance coil of `3 Omega`

`V =I xx R`

`=1.5 xx 3=4.5V`

Correct Answer is `=>` (C) `4.5 V`

Q 2316891779

If an input current 3 A flows through the circuit shown above, what is the value of the current flowing through the `4 Omega` resistor'?
NDA Paper 2 2007

(A)

1.6 A

(B)

0.8 A

(C)

0.75 A

(D)

0.4 A

Correct Answer is `=>` (D) 0.4 A

Q 2316791679

A wire of resistance `16 Omega` is bent in the form of a circle. What is the effective resistance between diametrically opposite points'?
NDA Paper 2 2007

(A)

`1 Omega`

(B)

`2 Omega`

(C)

`4 Omega`

(D)

`8 Omega`

Solution:

When wire is bent in the form of circle, each branch lias
resistance of `8 Omega` So, equivalent resistance between
diametrically opposite points, i.e., between `A` and `B` is

`1/(R_(AB))=1/8+1/8=1/4`

or `R_(AB)=4 Omega`

Correct Answer is `=>` (C) `4 Omega`

Kirchhoff's Laws and Wheat stone's Bridge

(i) `text(Junction law )- ` `sum I_text(junction) = 0`

(ii) `text( Loop law ) - ` `sum Delta V = 0 => sum E + RI = 0`

`Delta E = - sum RI`

`text(Wheat stone's Bridge )` `P/Q = R/S`

In this case `V_B = V_D` and the bridge will be balanced

Q 2356778674

The figure shows current in a part of electrical network. What is the value of current `I` ?
NDA Paper 2 2008

(A)

`0.2 A`

(B)

`0.1 A`

(C)

`0.3 A`

(D)

`0.5 A`

Solution:

According to Kirchhoff's law, the algebraic sum of current at a junction is zero.

According to question, At junction `A`

`3 + 1- I_1 = 0`

`=> I_1 = 4A`

At junction B

`4-2 -I_2 = 0`

`=>I_2 = 2 A`

At junction C

`2-I-1.5=0`

`=> I= 0.5` A

Correct Answer is `=>` (D) `0.5 A`

Heating Effect Of Electric Current

Heating Effect
`H = W/(4.2) = (VIt)/(4.2) = (V^2t)/(4.2R) = (I^2Rt)/(4.2)`

`text(Electrical Power)`

`P =VI = V^2/R = I^2R`

`text(Electrical Energy)`

The total work done by the source of emf is maintaining the electric current in the circuit for a given time is called electrical energy consumed in the circuit. Sl unit of electric energy is joule but another unit is watt-hour.

`text(Kilowatt-hour (kWh))`

It is bigger unit of electric energy. lt is known as Board of Trade Unit (BTO).

1 kWh = P(in kW) x t (in hour)
1 kWh = 1000 X 3600
1 kWh `= 3.6 xx 10^6` Joule

Q 2375812766

An electric lamp of `1 00 W` is used for `10 h` per day. The units of energy consumed in one day by the lamp is
NDA Paper 2 2011

(A)

`1` unit

(B)

`0.1` unit

(C)

`10` units

(D)

`100` units

Solution:

Energy consumed in a day `= 100 W xx 10 h`

`= 1000` Wh

`= 1 kWh = 1` unit

Correct Answer is `=>` (A) `1` unit

Q 2383480347

An electric heater is rated 1500 W, electric power costs Rs `2` per, kilo-watt-hour, then the cost of power for `10 h` running the heater is
NDA Paper 2 2013

(A)

` Rs 30`

(B)

`Rs 15`

(C)

`Rs 150`

(D)

Rs 25

Solution:

Power of electric heater `(p) = 1500 W`

`1` unit `= 1 kWh = 1. 5 xx 1`

`=1.5` unit

For `10` h,

The unit will be `= 1.5 xx 10 =15` unit

Cost of `1` unit `= Rs 2`

Cost of `15` unit `= Rs 15 xx 2 = Rs 30`

Correct Answer is `=>` (A) ` Rs 30`

Q 2334234152

A current `I` flows through a potential difference `V` in an electrical circuit containing a resistance `R`. The product of `V` and `I` i.e., `VI` may be understood as.
NDA Paper 2 2013

(A)

resistance `R`

(B)

heat generated by the circuit

(C)

thermal power radiated by the circuit

(D)

rate of change of resistance

Solution:

`P =VI`, it means that the thermal power is radiated by
the circuit.

Correct Answer is `=>` (C) thermal power radiated by the circuit

Q 2346578473

The rating of an electric lamp is `110 V` To use it on `220 V`, one will have to use which one of the following?
NDA Paper 2 2008

(A)

Transistor

(B)

Resistor

(C)

Transformer

(D)

Generator

Solution:

To use the electric lamp on `220 V`, one should use
resistor in series with the lamp.

Correct Answer is `=>` (B) Resistor

Q 2314267150

The power supply in India is at `220 V`, whereas that in the US is at `11 0 V` Which one among the following statements in this regard is correct?
NDA Paper 2 2012

(A)

`110 V` is safer but more expensive to maintain

(B)

`110 V` is safer and cheaper to maintain

(C)

`110 V` leads to lower power loss

(D)

`110 V` works better at higher latitudes

Solution:

Power loss is given by `P=VI` i.e., `PpropV` So `110 V` leads to lower power loss

Correct Answer is `=>` (C) `110 V` leads to lower power loss

Q 2315845769

If a heater coil is cut into two equal parts and only one part is used in the heater the heat generated will be
NDA Paper 2 2010

(A)

doubled

(B)

four times

(C)

one-fourth

(D)

halved

Solution:

Electrical power, `P=V^2/R`

`=> P prop 1/R`

But `R prop l`

so, `P prop 1/l`

If a heater coil is cut into two equal parts and only one part is used in heater, then heat generated will be double.

Correct Answer is `=>` (A) doubled

Q 2355391264

How many sixty watt ( `60 W`) bulbs may be safely used in a `240 V` supply with `4 A` fuse?
NDA Paper 2 2010

(A)

(B)

(C)

(D)

Solution:

Electric power, `P = V xx I`

`=240 xx 4=960 W`

`:.` Number of bulbs of `60 W = 960/60=16` bulbs.

Correct Answer is `=>` (D) 16

Q 2468191005

Two 220 V, 100 W bulbs are connected first
in series and then in parallel. Each time the
combination is connected to a 220V AC
supply line, the power drawn by the
combination in each case respectively will be
BCECE Stage 1 2013

(A)

50 W, 200W

(B)

50W, 20W

(C)

100W, 50W

(D)

200W, 150W

Solution:

In series, power, `P = V^2/(2R) = P/2 =50 W`

In parallel, power, `P = V^2/(R//2) = 2P = 200 W`

Correct Answer is `=>` (A) 50 W, 200W

Q 2375812766

An electric lamp of `1 00 W` is used for `10 h` per day. The units of energy consumed in one day by the lamp is
NDA Paper 2 2011

(A)

`1` unit

(B)

`0.1` unit

(C)

`10` units

(D)

`100` units

Solution:

Energy consumed in a day `= 100 W xx 10 h`

`= 1000` Wh

`= 1 kWh = 1` unit

Correct Answer is `=>` (A) `1` unit

Q 2346712673

A current `I` in a lamp varies with voltage `V` as shown in the figure given above. Which one of the following is the variation of power `P` with current `I` ?
NDA Paper 2 2009

Solution:

`P=V^2/R` or `P=(I^2R^2)/R=I^2R`

or `P prop i^2`

Hence, graph between `P` and `i` will be a hyperbola shown in fig (2)

Correct Answer is `=>` (B)

Q 2316712679

An electric iron of resistance `20 Omega` takes a current of `5 A`. The heat developed in joules in `30` s is
NDA Paper 2 2009

(A)

`5 kJ`

(B)

`10 kJ`

(C)

`15 kJ`

(D)

`20 kJ`

Solution:

Heat developed in an electric iron is

`H=i^2Rt`

Here, `i=5A,R=20 Omega,t=30s`

`:. H=(5)^2 xx 20 xx 20 xx 30=25 xx 20 xx 30`

`=15000 J=15 kJ`

Correct Answer is `=>` (C) `15 kJ`

Q 2386291177

Which one of the following is correct?
One unit of electric power is consumed when
NDA Paper 2 2007

(A)

1 A of current flows for 1 s at 220 V

(B)

1 A of current flows for 1 sat 1 V

(C)

100 A of current flows for 1 sat 10 V

(D)

10 A of current flows for 1 h at 100 V

Solution:

One unit of electric power `= 1 kWh = 1000` Wh

When `10 A` current flows for 1 h at `100 V`

Then, electricity consumed `= 10 xx 100 xx 1 Wh`

`= 1000 Wh`

`= 1 kWh`

Correct Answer is `=>` (D) 10 A of current flows for 1 h at 100 V

Electric cell

`text(Electromotive Force)` The potential difference across the terminals of the cell, when no current is being taken from the cell is called emf of a cell.

`E =W/q`

`text(Internal Resistance of a Cell)`
• If current i is being taken from the cell and potential difference across the cell is V, then internal resistance,

`r = (E -V)/i` or ` r = R (E/V - 1)`

`text(Relation between Electromotive Force, Terminal Voltage and Internal Resistance of Cell)`

The relation between electromotive fore `(E),` internal resistance ` (r)` and external resistance `(R.)` of circuit is

`E = V + ir`.

Galvanometer

•`text(Galvanometer)` Resistance of shunt, `S = (i_g G)/(i - i_g)`

where `G =` resistance of galvanometer,

`i =`range of ammeter

and `i_g =` safe current for galvanometer coil

To convert galvanometer into voltmeter a high resistance `(R)` is connected in series with the galvanometer, `R = V/i_g -G`

where , `V =` range of voltmeter