Tag: effects of electric current

Questions Related to effects of electric current

When we pay for our electricity bill, we are paying for the:

  1. charge used.

  2. current used.

  3. power used.

  4. energy used.


Correct Option: D
Explanation:

Electricity bill is measured in units where 1 unit is equal to kWh which is the unit of energy. Hence, when we pay for our electricity bill, we pay for the amount of electrical energy consumed by us.

An electric kettle is rated as $2.5\ kW, \ 250\ V$. Find the cost running the kettle for two hours at $60\ \text{paisa}/\text{unit}$.

  1. $Rs\ 6$

  2. $Rs\ 3$

  3. $Rs\ 9$

  4. $Rs\ 1$


Correct Option: B
Explanation:

Energy consumed, $E =2.5 \times 2 = 5\ kWh $
Cost, $C = 5 \times 0.60 = Rs\ 3 $

An electric bulb rated for $500$W at $100$ V is used in a circuit having a $200$V supply. The resistance R that one must put in series with the bulb, so that the power delivered in the bulb is $500$W is ______$\Omega$.

  1. $30$.

  2. $20$.

  3. $40$.

  4. $60$.


Correct Option: B

Four bulbs, each of rating (100 W, 220 V) and connected in parallel across a voltage supply of 220 V, are operated for five hours daily. If all the bulbs are replaced by LEDs of rating (8 W, 220 V), how many units of electrical energy will be saved every month (30 days)? 

  1. 55.2 units

  2. 60 units

  3. 4.8 units

  4. 32 units


Correct Option: A
Explanation:

For bulbs :

$Energy\,\,=\,power\times time$

Total hours in 30 days $5\times 30=150hours$

So, $ energy=0.1\times 150=15J $

 $  $Power of 4 bulbs

 $ =4\times 15=60kW $

 $  $For LED:

Total hours in 30 days $=150hours$

$E=0.008\times 150=1.2J$

So, power of 4 LEDs $=1.2\times 4=4.8hours$

Saved energy 

  $ =\,\,60-4.8 $

 $ =55.2units $


How much energy in kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

  1. 1500

  2. 5000

  3. 15

  4. 150


Correct Option: D
Explanation:

Energy consumed $=10\times 50\times 10\times 30\times 3600 J$
$[1 Kwh=3600\times 1000J]$
$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$


A geyser of $2.5 kw$ is used for $8$ hours daily. Calculate the monthly consumption ( 30 days) of electrical energy units. Also calculate the cost of electricity units consumed in a month if rate per unit is $3.50$

  1. $Rs. 2100.00$

  2. $Rs. 155.00$

  3. $Rs. 150.00$

  4. $Rs. 30.00$


Correct Option: A
Explanation:

Daily usage of electricity   $E _1 = 2.5 \ kw\times 8 = 20 \ kwh = 20 \ units$

Energy used in one month   $E = 30E _1 = 30\times 20 = 600\ units$
Cost    $ = Rs. 600\times 3.5 = Rs 2100.00$

If voltage across a bulb rated $220V-100W$ drops by $2.5$% of its rated value, the percentage of the rated value by which the power would decrease is

  1. $5$%

  2. $10$%

  3. $20$%

  4. $2.5$%


Correct Option: A

A padcular ohmmeter uses a battery to provide a potential difference across an unknown resistance  whose value S to be measured. The meter measures the resulting current through this resistor and is calibrated to read out corresponding value of resistance. Suppose that this ohmmeter is used to measure he resistance of a typical incandescent tungsten-filament light bulb. The value of the resistance of the light bulb will be

  1. less then when the bulb will be in use in a 120 volt circuit

  2. more then when the bulb will be in use in a 120 volt circuit

  3. the same as then when the bulb will be in use in a 120 volt circuit

  4. more information when needed to determine whether it's A,B and C


Correct Option: B

A $500\ W$ heating unit is designed to operate on a $115\ V$ line. If line voltage drops to $110\ V$ line, the percentage drop in heat output will be:

  1. $7.6\ \%$

  2. $8.5\ \%$

  3. $8.1\ \%$

  4. $10.2\ \%$


Correct Option: B
Explanation:

Given:
$H _1 = 500\ W$
$V _1 = 115\ V$
$V _2 = 110\ V$

From Joule's Law of heating,
$H _1 = \cfrac{(V _1)^2}{R}$ and $H _2 = \cfrac{(V _2)^2}{R}$
$\Rightarrow R = \cfrac{(V _1)^2}{H _1} = \cfrac{(V _2)^2}{H _2}$
$\Rightarrow H _2 = \cfrac{V _2^2}{V _1^2} H _1$
$\therefore H _2 = \cfrac{(110)^2}{(115)^2} (500)$
$\therefore H _2 = 457.46 W$

The percentage drop in heat output will be:
$\cfrac{H _2-H _1}{H _1} \times 100 = \cfrac{500 - 457.46}{500} \times 100 = \cfrac{42.54}{500} \times 100 = 8.5\ \%$

An electric heater operating at $220\ V$ boils $5\ l$ of water in $5\ \text{minutes}$. If it is used on a $110\ V$ line, it will boil the same amount of water in:

  1. $10\ \text{minutes}$

  2. $20\ \text{minutes}$

  3. $5\ \text{minutes}$

  4. $1\ \text{minute}$


Correct Option: B
Explanation:
Heat required to boil the water will be same in the two cases.
Heat produced by heater is:
$Q = \cfrac{V^2}{R} t$
$V^2 = \cfrac{QR}{t}$
$V^2 \propto \dfrac{1}{t}$

$\cfrac{t _2}{t _1} = \cfrac{V _1^2}{V _2^2}$
$t _2 = \dfrac{220^2 \times 5}{110^2}$
$t _2 = 20\ min$