Question

Solve Fully and send full answer with calculations....
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Answer 1

Given

• There are four 4Ω resistors
• Potential Difference = 2V

To Find

• Current when the key is open
• Current when the key is closed

Solution

● V = IR [Ohm's law]

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

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

✭ Series Connection :

• Both these connection are separated into two series and parallel connection. First two 4 Ω resistance in series and these are connected in parallel

→ R(eq) = R₁ + R₂ ..... + Rₙ

→ R(eq) = 4 + 4

→ R(eq) = 8 Ω

✭ Parallel Connection :

• R₁ = 8Ω
• R₂ = 8Ω

→ 1/Rₙₑₜ = 1/R₁ + 1/R₂ ..... + 1/Rₙ

→ 1/Rₙₑₜ = 1/8 + 1/8

→ 1/Rₙₑₜ = (1+1)/8

→ 1/Rₙₑₜ = 2/8

→ Rₙₑₜ = 8/2

→ Rₙₑₜ = 4Ω

✭ Current when the key is closed :

→ V = IR

→ 2 = I × 4

→ 2/4 = I

→ Current = 0.5 A

━━━━━━━━━━━━━

• When the key is open then there would be no current, that is Current = 0 A

Answer 2

☆Answer☆

Given,

There are four resistors of 4 ohm each.

Potential Difference, V = 2V

In the circuit, two resistors are connected in series.

R equivalent = R1 + R2

=> (4+4) ohm

=> 8 ohm

In Parallel Connection:

1/(R equivalent) = 1/R1 + 1/R2

=> (1/8+1/8) ohm

=> 2/8 ohm

R = (8/2)

R equivalent = 4 ohm

Now,

V = IR

2 = I×4

I = 2/4

=> 0.5 Amp

When the circuit open there would be no current and when circuit is closed current will be 0.5 Ampere.

✔