Three resistors of resistance 6 Ohm each are provided. Explain with circuit diagram so that a
total resistance of a) 4 Ohm, b) 9 Ohm is obtained by connecting all the three resistors
Answers
Solution :
As per the given data ,
- R = R 1 = R 2 = R 3 = 6 Ω
( 1 ) 4 ohm
Step I :
Two 6 Ω resistors are connected in series with each other .
Equivalent resistance for resistors connected in series is given by ,
→ Rs = R 1 + R 2 ... + R n
Hence ,
→ Rs = R + R
→ Rs = 6 + 6
→ Rs = 12 Ω
Step II :
Rs is now connected in parallel with the third 6 Ω resistor .
Equivalent resistance for resistors connected in parallel is given by ,
→ 1 / Rp = 1 / R 1 + 1 / R 2 ... + 1 / Rn
Hence ,
→ 1 / Rp = 1 / Rs+ 1 / R
→ 1 / Rp = 1 / 12 + 1 / 6
→ 1 / Rp = 1 + 2 / 12
→ 1 / Rp = 3 / 12
→ Rp = 4Ω
Hence the net equivalent resistance of this combination is 4 Ω .
(2) 9 ohm
Step I :
Two 6 Ω resistors are connected in parallel with each other .
Equivalent resistance for resistors connected in parallel is given by ,
→ 1 / Rp = 1 / R 1 + 1 / R 2 ... + 1 / Rn
Hence ,
→ 1 / Rp = 1 / R+ 1 / R
→ 1 / Rp = 1 / 6 + 1 / 6
→ 1 / Rp = 2 / 6
→ Rp = 3 Ω
Step II :
Rs is now connected in series with the third 6 Ω resistor .
Equivalent resistance for resistors connected in series is given by ,
→ Rs = R 1 + R 2 ... + R n
Hence ,
→ Rs = Rp + R
→ Rs = 3 + 6
→ Rs = 9 Ω
The equivalent resistance of the combination is 9Ω