Tag: physics

Questions Related to physics

Statement A: Step up transformer converts low voltage, high current to high voltage, low current.
Statement B: Transformer works on both ac and dc.

physics electro-magnetism applications of electromagnets transformer transformers

  1. A is true but B is false

  2. Both A and B are true

  3. A is false but B is true

  4. Both A and B are false

Show answer
Correct Option: A
Explanation:

Step up transformer increases the voltage in secondary coil and decreases current in secondary coil. Transformer works only on ac current because in DC current, there is no change of flux associated with the coil

Maximum efficiency of a transformer depends on :

physics electro-magnetism applications of electromagnets transformer transformers

  1. the working conditions of technicians

  2. whether copper loss $=$1/2 x iron loss

  3. whether copper loss $=$ iron loss

  4. whether copper loss $=$2 x iron loss

Show answer
Correct Option: C
Explanation:

The efficiency of the transformer is defined as ratio of useful power output to the input power. The two being measured is the same unit. Its unit is either watts or kw. Transformer efficiency is denoted by $\eta$

$\eta = \dfrac{\text{output power}}{\text{input power}} = =\dfrac{\text{output power}}{\text{input power + losses}}$

$\eta  = \dfrac{\text{output power}}{\text{input power + iron losses of copper + losses}}$

$\eta = \dfrac{V _2I _2 \cos \phi _2}{V _2 I _2 \cos \phi _2 + P _i + P _c}$

Where

$V _2  = $ secondary terminal voltage

$I _2 = $ Full load secondary current.

$\cos \phi _2 = $ Power factor of the load

$P _i = $ Iron losses = constant loss = exactly current loss + hysteresis loss

$P _c = $ full load copper loss

The efficiency is a function of load i.e. lead current $I _2$ assuming $\cos \phi$ constant. The secondary terminal voltage $V _2$ is also assumed constant.

$\dfrac{d \eta}{d I _2} = 0$

Now $\eta = \dfrac{V _2I _2\cos \phi _2}{ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} }$

$\dfrac{d \eta}{d I _2} = \dfrac{d}{d I _2}(\dfrac{V _2I _2\cos \phi _2}{ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} }) = 0$

$( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e}) \dfrac{d}{d I _2}( V _2I _2\cos \phi _2) – (V _2I _2\cos \phi _2) \dfrac{d}{d I _2}( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e}) = 0$

$( V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e})(V _2 \cos \phi _2) – (V _2 I _2 \cos \phi _2)(V _2 \cos \phi _2 + 2 I _2 R _{2e}) = 0$

Cancelling $V _2 \cos \phi _ 2$ from both the terms we get

$ V _2I _2\cos \phi _2 + P _i + I _2^2 R _{2e} – V _2I _2\cos \phi _2 – 2I^2 R _{2e} = 0 $

$P _i – I^2 R _{2e} = 0$

$P _i = I^2 R _{2e} = P _{Cu}$

$P _i = P _{Cu}$

 So, condition to achieve maximum efficiency is that

Copper Loss = Iron Loss

A 220 - volt input is supplied to a Transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is

physics electro-magnetism applications of electromagnets transformer transformers

  1. 5.0 ampere

  2. 3.6 ampere

  3. 2.8 ampere

  4. 2.5 ampere

Show answer
Correct Option: A
Explanation:

$\mu =\frac { { V } _{ s }{ I } _{ s } }{ { V } _{ p }{ I } _{ p } } =0.8\Rightarrow { I } _{ p }=\frac { { V } _{ s }{ I } _{ s } }{ { V } _{ p }\mu  } =\frac { 440\times 2 }{ 220\times .8 } =5A$

Electrical energy is transmitted over large distance at high alternating voltages Which of the following statements is (are) correct?

physics electro-magnetism applications of electromagnets transformer transformers

  1. for a given power level there is a lower current

  2. lower current implies less power loss

  3. transmission lines can be made thinner

  4. It is easy to reduce the voltage at the receiving end using step-down transformers

Show answer
Correct Option: A,B,D
Explanation:

Power = Voltage x Current
As voltages are high so currents are lower. This is done to reduce the resistive losses.
Resistance Loss = ${ { I } _{ rms }^{ 2 } }R$
So, lower current implies less power loss. 
As resistance is inversely proportional to their diameter; hence decreasing their diameter by x will increase resistance by $x^2$ and hence increasing power loss hence not done.
As the voltage required for domestic and commercial uses are pretty less then they can be easily brought down by step-down transformers. 

A direct current (DC) motor is connected to a battery by means of two leads. The motor moves for slightly lessthan half the cycle and comes to halt. Which of the following components is missing?

physics electro-magnetism applications of electromagnets transformer transformers

  1. one of the brushes

  2. commutator

  3. slip ring

  4. one of the two magnets

Show answer
Correct Option: C
Explanation:

Answer is C.

Slip-ring commutator merely maintains a connection between the moving rotor and the stationary stator.
Slip rings are useful for DC motors where the current has to change direction every half revolution whereas split rings are used to make constant current connection to a commutator, which is what we want for an AC motor.
Hence, if the slip rings are not present, then the armature stops its rotation.

Tum ratio in a step up transformer is 1 : 2 if a Lechlanche cell of 1.5 V is connected across the input, then the voltage across the output will be 

physics electro-magnetism applications of electromagnets transformer transformers

  1. 0.1 V

  2. 1.5 V

  3. 0.75 V

  4. Zero

Show answer
Correct Option: D
Explanation:

A transformer can not step up a d.c. input so output potential here will be zero. No potential will be induced in the secondary coil.

Large transformers, when used for some time, become very hot and are cooled by circulating oil. The heating of the transformer is due to

physics electro-magnetism applications of electromagnets transformer transformers

  1. The heating effect of current alone

  2. Hysteresis loss alone

  3. Both the heating effect of current and hysteresis loss

  4. Intense sunlight at noon

Show answer
Correct Option: C
Explanation: 

Heat loss in a magnetic circuit is due to two reasons one is hysteresis and  other is due to ohmic loss due to eddy current .Laminations and circulatingoil are some of methods to avoid heating. Hysteresis loss is entirelymaterial property while eddy current loss depends on geometry of core and amount of current.

A certain transformer has a turns ratio of 1 and a 0.85 coefficient of coupling. When 2 V ac is applied to the primary, the secondary voltage is

physics electro-magnetism applications of electromagnets transformer transformers

  1. 1.7 V

  2. 0.85 V

  3. 1 V

  4. 0 V

Show answer
Correct Option: A
Explanation:

Coefficient coupling  $\eta = 0.85$
Input voltage in primary coil   $V _p = 2 \ V$
Turn ratio  $\dfrac{N _s}{N _p} = 1$
Thus output voltage in secondary coil   $V _s = \eta \dfrac{N _s}{N _p}V _p$
$\implies \ V _s = 0.85\times 1\times 2 = 1.7 \ V$

To step 120 V ac up to 900 V ac, the turns ratio must be

physics electro-magnetism applications of electromagnets transformer transformers

  1. 75

  2. 750

  3. 7.5

  4. 0.13

Show answer
Correct Option: C
Explanation:

Input voltage in primary coil  $V _p = 120 \ V$
Output voltage in secondary coil   $V _s = 900 \ V$
Turn ratio    $  \dfrac{N _s}{N _p} = \dfrac{V _s}{V _p} = \dfrac{900}{120} = 7.5$

The primary winding of a power transformer should always be

physics electro-magnetism applications of electromagnets transformer transformers

  1. open

  2. shorted

  3. switched

  4. fused

Show answer
Correct Option: D
Explanation:

When the secondary load is overloaded, it draws maximum current in the primary side due to which damaging the primary side as well as appliances. Therefore primary winding of power transformer is fused. Suppose peak voltage of power source is varying then current will also fluctuate so also to limit the current it is fused.