Question

seeing this diagram find out the value of I?

Answer 1

Given :

Four resistors are connected as shown in the figure.

Voltage of battery = 6V

To Find :

Net current flow in circuit.

Solution :

• R₁ = 2 Ω
• R₂ = 6 Ω
• R₃ = 1.5 Ω
• R₄ = 3 Ω

First of all we need to simplify the given circuit. See the attached image for better understanding.

From the attachment, it is clear that R₁ and R₂ are connected in parallel.

So their equivalent resistance will be;

➠ 1/R₁₂ = 1/R₁ + 1/R₂

➠ 1/R₁₂ = 1/2 + 1/6

➠ R₁₂ = 6/4

➠ R₁₂ = 1.5 Ω

After solving first part, R₁₂ and R₃ come in series$.$

➠ R₁₂₃ = R₁₂ + R₃

➠ R₁₂₃ = 1.5 + 1.5

➠ R₁₂₃ = 3 Ω

Finally, R₁₂₃ and R₄ come in parallel. So net equivalent resistance of circuit will be

➠ 1/R = 1/R₁₂₃ + 1/R₄

➠ 1/R = 1/3 + 1/3

➠ 1/R = 2/3

➠ R = 1.5 Ω

∴ Net equivalent resistance = 1.5 Ω

Answer 2

Given :

Four resistors are connected as shown in the figure.

Voltage of battery = 6V

To Find :

Net current flow in circuit.

Solution :

R₁ = 2 Ω

R₂ = 6 Ω

R₃ = 1.5 Ω

R₄ = 3 Ω

First of all we need to simplify the given circuit. See the attached image for better understanding.

From the attachment, it is clear that R₁ and R₂ are connected in parallel.

So their equivalent resistance will be;

➠ 1/R₁₂ = 1/R₁ + 1/R₂

➠ 1/R₁₂ = 1/2 + 1/6

➠ R₁₂ = 6/4

➠ R₁₂ = 1.5 Ω

After solving first part, R₁₂ and R₃ come in series..

➠ R₁₂₃ = R₁₂ + R₃

➠ R₁₂₃ = 1.5 + 1.5

➠ R₁₂₃ = 3 Ω

Finally, R₁₂₃ and R₄ come in parallel. So net equivalent resistance of circuit will be

➠ 1/R = 1/R₁₂₃ + 1/R₄

➠ 1/R = 1/3 + 1/3

➠ 1/R = 2/3

➠ R = 1.5 Ω

∴ Net equivalent resistance = 1.5 Ω