# 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

### Factors affecting resistance

1. The conductor’s cross-sectional area
2. the conductor’s length
3. the conductor’s composition
4. The conducting material’s temperature

### Applications of resistor

1. When accurate measurement, high sensitivity, and balanced current regulation are needed, wire wound resistors are used in applications like shunts with ampere meters.
2. Flame detectors, burglar alarms, photographic equipment, etc. all use photoresistors.
3. Temperature and voltmeter control is accomplished by resistors.
4. Amplifiers, telephony, oscillators, and digital multimeters all employ resistors.
5. 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.

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