Establish a relationship to determine the equivalent resistance R of a combination of three resistors having resistances R _{ 1 } , R _{ 2 } and R _{ 3 } connected in series. Calculate the equivalent resistance of the combination of three resistors of 2 Ω, 3 Ω and 6 Ω joined in parallel.
Answer:
(i)
An applied potential V produces current I in the resistors R1, R2 and R3, causing a potential drop V1, V2 and V3 respectively through each resistor.
Total Potential, V = V _{ 1 } + V _{ 2 } + V _{ 3 }
Let R be the equivalent resistance.
According to ohm’s law,
V = IR
Therefore,
- V 1 = IR _{ 1 }
- V 2 = IR _{ 2 }
- V 3 = IR _{ 3 }
Thus, V = IR _{ 1 } + IR _{ 2 } + IR _{ 3 }
IR = I(R _{ 1 } + R _{ 2 } + R _{ 3 } )
R = R _{ 1 } + R _{ 2 } + R _{ 3 }
This proves that equivalent Resistance R is the sum of resistances R _{ 1 } , R _{ 2 } , R _{ 3 }
(ii)
Three resistors 2 Ω, 3 Ω and 6 Ω, are connected in parallel combination.
Let R be the equivalent resistance.
1 / R = 1 / 2 + 1 / 3 + 1 / 6
= ( 3 + 2 + 1 ) / 6
= 6 / 6
Therefore, R = 1 ohm
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