Question

3 resister resistences 3ohm,6ohm,9ohm are
Connected to series 15v batter Calculate
the efective resistance and the current flowing to
the sercuit?​

Answer 1

Explanation:

R1=3ohm R2=4ohm,R3=6ohm

As the resistors are connected in parallel

1/Rtotal=1/3+1/4+1/6

1/Rtotal=(4+3+2)/12

1/Rtotal=9/12

Rtotal=12/9

Rtotal=4/3 ohm

Total current=Voltage /Rtotal

Itotal=2/(4/3)

Itotal=6/4

Itotal=3/2 Ampere

For resistance in the parallel ,voltage across each resistor is same and total current is sum of current through each resistor.

Current through R1=V/R1=2/3 Ampere

Current through R1=V/R2=2/4=1/2Ampere

Current through R1=V/R3=2/6=1/3Ampere

Total current is 3/2 Ampere

Answer 2

Answer :

Three identical resistors of 3Ω, 6Ω and 9Ω are connected in series with a battery of 15V.

We have to find effective resistance and net current flow in the circuit$.$

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◈ Equivalent resistance :

Equivalent resistance of series connection is given by

• Rs = R₁ + R₂ + R₃ + ... + Rn

➝ Rs = R₁ + R₂ + R₃

➝ Rs = 3 + 6 + 9

➝ Rs = 18Ω

◈ Net current flow :

⇒ V = I × Rs

⇒ 15 = I × 18

⇒ I = 15/18

⇒ I = 0.83A