Tag: physics

Questions Related to physics

Household electrical appliances are joined using .......... combination of resistors.

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  1. Parallel

  2. Alternating

  3. Continuous

  4. Series

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Correct Option: A
Explanation:

Household electrical appliances are always joined in parallel.

The current through a 60 W lamp rated for 250 V is 

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  1. 0.24 A

  2. 4.2 A

  3. 0.5 A

  4. 6 A

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Correct Option: A
Explanation:
The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. The power is given by the product of applied voltage and the electric current. That is, $P=VI$. 

For a 60 W lamp rated for 250 V, the current through is given as $I=\dfrac { P }{ V } =\dfrac { 60 }{ 250 } =0.24A$.

Hence, the current flowing through the lamp is 0.24 A.

Substituting I=V/R in the above formula, we get, $P=\dfrac { { V }^{ 2 } }{ R } $.

Given that the voltage is 250 V and the power is 60 W, the resistance of the 

bulb is calculated as follows.

$R=\dfrac { { V }^{ 2 } }{ P } =\dfrac { { 250 }^{ 2 } }{ 60 } =1041.67\Omega$.

When the voltage drops to 200 V, the power is calculated as follows.

$P=\dfrac { { V }^{ 2 } }{ R } =\dfrac { { 200 }^{ 2 } }{ 1041.67 } =\quad 38.40W$

Hence, the power of the bulb is reduced to 38.40 W.

We should not use many electrical appliances simultaneously. True/False?

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  1. True

  2. False

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Correct Option: A
Explanation:

When many appliances are switched on simultaneously, a lot of current flows through the main circuit and the current may exceed the permissible amount. This causes overloading and may cause a fire. Thus many electrical appliances should not be used simultaneously.

An electric heater consists of a nichrome coil and runs under $220 V$, consuming $1 kW$ power. Part of its coil burned out and it was reconnected after cutting off the burnt portion, The power it will consume now is:

physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

  1. More than $1 kW$

  2. Less that $1 kW$, but not zero

  3. $1 kW$

  4. $0kW$

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Correct Option: A
Explanation:

Answer is A.

The electrical resistance of a wire would be expected to be greater for a longer wire, less for a wire of larger cross sectional area, and would be expected to depend upon the material out of which the wire is made.The resistance of a wire can be expressed as $R=\rho \frac { L }{ A } $, 
where,
$\rho $ - Resistivity  - the factor in the resistance which takes into account the nature of the material is the resistivity
L - Length of the conductor
A - Area of cross section of the conductor.
From this relation, we observe that the length is directly proportional to the resistance and the area of cross section is inversely proportional to the resistance.
In this case, the length of the nichrome coil is reduced due to the burn and so the resistance will also be reduced proportionally. 
When the resistance is decreased, more current flows through the coil and apparently, more power is consumed by the heater as power consumed P = VI.
Hence, the power it will consume now is more than 1 kW.

An air conditioner is rated $240\;V,\;1.5\;kW$. The air conditioner is switched on for $8$ hours each day. How much electrical energy is consumed in $30$ days ?

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  1. $360\;kW\;h$

  2. $8.64\;kW\;h$

  3. $120\;kW\,h$

  4. $240\;kW\;h$

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Correct Option: A
Explanation:

Answer is A.

If a certain amount of power is dissipated for a given time, then energy is dissipated. Energy (powertime) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.
Energy dissipated = Pt .
In this case, the power dissipated is 1.5 kW and it runs for 8 hours a day.
So, in a day, the energy dissipated is 1.5 * 8 = 12 Kwh.
For 30 days, the power dissipated is 12 * 30 = 360 Kwh.
Hence, the electrical energy is consumed by the air conditioner in 30 days is 360 Kwh.

A washing machine rated $300W$ is operated for one hour/day. If the cost of a unit is Rs $3.00$ then the cost of the energy to operate a washing machine for the month of March is

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  1. $Rs 25.60$

  2. $Rs 27.50$

  3. $Rs 27.90$

  4. $Rs 26.90$

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Correct Option: C
Explanation:

In this case, the washing machine that is rated $ 300\ W$ is operated for $1$ hr/day. The electric bill has to be calculated for the month of March. That is, the total number of hours the washing machine is operated in this month is $31$ hours as this month has $31$ days.
The total energy that is used by the washing machine in a day is given by the formula $Q=Pt$.
Here, the power is $300\ W$ and the time used is $1\ hour$.
So, $Q=300W\times 1h=300Wh= 0.3\ kWh$
Therefore, the total number of $kWh$ for a month is $0.3kWh\times 31=9.3\ kWh $
The cost of $1\ kWh$ is given as $Rs. 3$.

Hence, the total cost for a month is $=9.3kWh\times 3=Rs. 27.90$

A neon lamp is connected to a voltage a.c. source. The voltage is gradually increased from zero volt. It is observed that the neon flashes at $50 V$. The a.c, source is now replaced by a variable dc source and the experiment is repeated. The neon bulb will flash at ?

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  1. $50V$

  2. $70V$

  3. $100V$

  4. $35V$

Show answer
Correct Option: B
Explanation:

Answer is B.

The RMS (Root mean square) value tells us what equivalent DC voltage we would need to get the same power, for the neon bulb to glow.
In this case, the neon bulb glows at 50 V ac voltage. Therefore, for the bulb glow with dc voltage, $50\times \sqrt { 2 }  =50\times 1.414=70V$ dc voltage should be applied.
Hence, the neon bulb will flash at 70 V dc.

Two dissimilar bulbs are connected in series, which bulb will be brighter ?

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  1. low resistance bulb

  2. more resistance bulb

  3. high current pass bulb.

  4. low current pass bulb

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Correct Option: B
Explanation:

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component.
When two bulbs are connected in series the first bulb will receive more current. As the bulb has its own resistance, less current will flow to the next bulb.
Hence the first bulb will glow more.

The wire having a green plastic covering is a

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  1. live wire.

  2. neutral wire.

  3. earth wire.

  4. none of these.

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Correct Option: C
Explanation:

Black, Red and Blue are used for hot wires and White is used as the neutral wire in a 120/208 V circuit. Brown, Orange and Yellow are used as hot wires and gray is used as the neutral wire in a 277/480 V. For grounding, regardless of the voltage, Green is used.The wire having a green plastic covering is a earth wire.

An electrician not aware of the colour coding of resistors connected two resistors A and B in series to a 6V battery of internal resistance $3\Omega$ and an ammeter. The ammeter connected in the circuit was not working and hence he disconnected the ammeter from the circuit. The sequence of the colour bands on resistor. A is yellow, violet and brown while that on resistor B is red, violet and black respectively. By using the colour coding of resistors, help the electrician to determine the current flowing through the circuit. 

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  1. $12$ mA

  2. $24$ mA

  3. $48$ mA

  4. $32$ mA

Show answer
Correct Option: A
Explanation:

Given: resistors A and B connected in series to a $6V$ battery of $3\Omega$ internal resistance. The sequence of the colour bands on resistor. A is yellow, violet and brown while that on resistor B is red, violet and black respectively. 

To find the current flowing through the circuit.
Solution:
According to the question the A and B are 3-band resistors, so the thirdcode will be a multiplier.
And using the standard resistor color code table the value of:
yellow is 4, violet is 7, brown is $\times 10^1$, red is 2, black is $\times 10^0$.
The corresponding value of A and B are:
A= $47\times 10^1=470\Omega$, B= $27\times 10^0=27\Omega$
Now A and B are in series are also the internal resistance will also be in series. So the effective resistance of the circuit will be $R _{eq}=470+27+3=500\Omega$.
We know according to Ohm's Law,
$V=IR\implies I=\dfrac VR\\implies I=\dfrac 6{500}=0.012A$
Or the current flowing through the circuit = $12mA$