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How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of 1Ω?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The three resistors of resistances 2Ω, 3Ω and 6Ω have to be connected in parallel to give a total resistance of 1Ω. Steps to calculate:
Step 1: Given Data:
The given resistances are:
- R1 = 2Ω
- R2 = 3Ω
- R3 = 6Ω
Due to the equivalent resistance’s lower value than each resistor’s resistance, the resistors R1, R2 and R3 must be connected in parallel.
Step 2: Formula used:
Therefore, their equivalent resistance:
1/Req = 1/R1 + 1/R2 + 1/R3
Step 3: Calculate the equivalent resistance:
1/Req = ½ + ⅓ + ⅙
= (3 + 2 + 1)/6
= 6/6
= 1
Req = 1Ω
Table of content
Factors affecting resistance
- The conductor’s cross-sectional area
- the conductor’s length
- the conductor’s composition
- The conducting material’s temperature
Applications of resistor
- When accurate measurement, high sensitivity, and balanced current regulation are needed, wire wound resistors are used in applications like shunts with ampere meters.
- Flame detectors, burglar alarms, photographic equipment, etc. all use photoresistors.
- Temperature and voltmeter control is accomplished by resistors.
- Amplifiers, telephony, oscillators, and digital multimeters all employ resistors.
- Additionally, they are utilized in transmitters, demodulators, and modulators.
Therefore, three resistors of resistances 2Ω, 3Ω and 6Ω have to be connected in parallel to give a total resistance of 1Ω
Summary:
How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of 1Ω?
To get a total resistance of 1Ω, the three resistors with resistances of 2Ω, 3Ω and 6Ω must be linked in parallel. When current flows through a light bulb or conductor, the conductor presents some obstacle to the current. This impairment is called electrical resistance and is represented by R.