Derive expression for resistors connected in Parallel. Draw Diagram.Three resistors of 20
Ohms, 4 Ohms and 5 Ohms are connected in parallel in a with the battery of 6 V. Find Total
Current.
Answers
Solutions:
(1)
Consider , three resistors R 1 , R 2 and R 3 are connected in parallel combination with each other .
Let the voltage of the battery connected be V and the total current flowing through the circuit be I .
In parallel combination the current flowing through each resistor is different but the voltage across each resistor remains the same .
Hence , let the current flowing through R 1 , R 2 and R 3 be I 1 , I 2 and I 3 respectively ,
Net current flowing ,
→ I = I 1 + I 2 + I 3
We know that ,
→ I = V / R [ By ohm's law ]
Hence ,
→ V / R = V / R 1 + V / R 2 + V / R 2
→ 1 / R = 1 / R 1 + 1 / R 2 + 1 / R 3
For n no of resistors connected in parallel ,
→ 1 / R = 1 / R 1 + 1 / R 2 + 1 / R 3 .... + 1 / Rn
(2)
Given :
- R 1 = 20 Ω
- R 2 = 4 Ω
- R 3 = 5 Ω
- V = 6 V
Solution :
All the resistors are connected in parallel to each other .
The Equivalent resistance of resistors connected in parallel is given by ,
→ 1 / R = 1 / R 1 + 1 / R 2 + 1 / R 3
→ 1 / R = 1 / 20 + 1 / 4 + 1 / 5
→ 1 / R = 1 + 5 + 4 / 20
→ 1 / R = 10 / 20
→ R = 2 Ω
The net equivalent resistance of the combination is 2 Ω .
By ohm's law ,
→ I = V / R
→ I = 6 / 2
→ I = 3 A
The net current flowing through the circuit is 3 A .