Question

Please give an answer to the following question in the picture (attached) ​

Answer 1

Given

• There are four 10Ω resistors
• Potential Difference = 3V

To Find

• Current in the circuit

Solution

☯ V = IR [Ohm's law]

☯ Rₙₑₜ = R₁ + R₂ .... + Rₙ [Series Connection]

☯ 1/Rₙₑₜ = 1/R₁ + 1/R₂ .... + 1/Rₙ [Parallel Connection]

✭ According to the Question :

• So here we have the top three resistors connected in series and this combination is connected in parallel to the fourth resistor

➞ R(eq) = R₁ + R₂ .... + Rₙ

➞ R(eq) = 10 + 10 + 10

➞ R(eq) = 30 Ω

So for parallel connection,

• R₁ = 30Ω
• R₂ = 10Ω

➞ 1/Rₙₑₜ = 1/R₁ + 1/R₂ .... + 1/Rₙ

➞ 1/Rₙₑₜ = 1/30 + 1/10

➞ 1/Rₙₑₜ = (1+3)/30

➞ 1/Rₙₑₜ = 4/30

➞ Rₙₑₜ = 30/4

➞ Rₙₑₜ = 7.5Ω

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✭ Current in the circuit :

➞ V = IR

➞ 3 = I × 7.5

➞ 3/7.5 = I

➞ Current = 0.4 A

∴ The current in the circuit is 0.4 A

Answer 2

V= 3 V(given)

I=?

R(total resistance)=?

If the diagram is simplified it will look as shown below,

So, R1=10 Ω

R2= 10 Ω + 10 Ω + 10 Ω

R2=30 Ω

R1 and R2 are in parallel

1/R=1/R1 + 1/R2

1/R=1/10 + 1/30

R=30/4

R=7.5 Ω

V=IR

3=I X 7.5

I=3/7.5

I=2/5 A

I=0.4 A

Hence, 0.4 A current is drawn from battery.

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