
If 1 unit of electricity costs $0.20$ , how much does it cost to switch on a heater marked 120V, 3A for 90 minutes ?
A) 0.11
B) 2.70
C) 64.80
D) 180.00
Answer
125.4k+ views
Hint: The cost of an electronic Appliance depends on the amount of energy used by it. The commercial unit of energy is kWh. Thus the cost to switch on the heater can be calculated if we know the value of energy used by it in commercial units, which is kWh. Then we can find the cost by multiplying the cost of 1 electricity unit and the total kWh of energy consumed.
Formula used:
$Energy = Power \times Time$
$Power = Voltage \times Current$
Complete step by step solution:
Let V be the voltage across the heater
Voltage across the heater $V = 120$ volts
Converting it into kilovolts (therefore dividing it by 1000, as $1kV = 1000$ volts)
Thus, $V=0.12kV$
Let I be the current in the heater
Current flowing through the heater $I = 3A$
Formula for power of the heater is given by $P = VI$
By substituting the values, we get the value of power of the heater,
\[\therefore \;P = 0.12 \times 3 = 0.36\;kW\]
Let T be the time for which heater is used
Time of usage $(T) = 90{\text{ minutes}}$
Converting it into hours (therefore dividing it by $60$ , as \[1hr = 60mins\] )
Thus time of usage of heater\[ = 1.5{\text{ }}hrs\]
Thus energy consumed is given by $E = PT$
Substituting values,
\[E = 0.36 \times 1.5 = 0.54\;kWh\]
Now the total electricity units consumed by the heater are \[0.54\] kWh.
And cost of 1 unit of electricity \[ = 0.20\]
Cost to switch on heater for 90 minutes \[ = {\text{ }}\;0.54 \times 0.2\] \[ = 0.11\]
Hence, the right answer is (A) that is \[0.11\]
Note: The cost of using any electronic appliance can also be found by using a shortcut formula, that is
Cost \[ = {\text{ }}\dfrac{{\left( {V \times I \times T \times C} \right)}}{{1000}}\] where $V$ voltage in volts, $I$ is current in ampere, $T$ is time in hours and $C$ is cost of one electricity unit.
Formula used:
$Energy = Power \times Time$
$Power = Voltage \times Current$
Complete step by step solution:
Let V be the voltage across the heater
Voltage across the heater $V = 120$ volts
Converting it into kilovolts (therefore dividing it by 1000, as $1kV = 1000$ volts)
Thus, $V=0.12kV$
Let I be the current in the heater
Current flowing through the heater $I = 3A$
Formula for power of the heater is given by $P = VI$
By substituting the values, we get the value of power of the heater,
\[\therefore \;P = 0.12 \times 3 = 0.36\;kW\]
Let T be the time for which heater is used
Time of usage $(T) = 90{\text{ minutes}}$
Converting it into hours (therefore dividing it by $60$ , as \[1hr = 60mins\] )
Thus time of usage of heater\[ = 1.5{\text{ }}hrs\]
Thus energy consumed is given by $E = PT$
Substituting values,
\[E = 0.36 \times 1.5 = 0.54\;kWh\]
Now the total electricity units consumed by the heater are \[0.54\] kWh.
And cost of 1 unit of electricity \[ = 0.20\]
Cost to switch on heater for 90 minutes \[ = {\text{ }}\;0.54 \times 0.2\] \[ = 0.11\]
Hence, the right answer is (A) that is \[0.11\]
Note: The cost of using any electronic appliance can also be found by using a shortcut formula, that is
Cost \[ = {\text{ }}\dfrac{{\left( {V \times I \times T \times C} \right)}}{{1000}}\] where $V$ voltage in volts, $I$ is current in ampere, $T$ is time in hours and $C$ is cost of one electricity unit.
Recently Updated Pages
JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Main Login 2045: Step-by-Step Instructions and Details

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
