Question

Please give full working also.

Answer 1

Question:

An electrical circuit is shown in the figure. R₁ , R₂ , R₃ and R₄ are resistances. The voltage drop in R₁, R₂ & R₄ are 2, 5 and 15 V respectively. The value of R₃ = 200 Ω. The value of R₁ in the circuit is

Answer:

R₁ = 50 Ω

Explanation:

In series combination,

• The same current flows through each resistor.
• The voltage across the combination is the sum of the individual potential differences across each resistor.

In parallel combination,

• The voltage is same across the resistors.
• The current flowing through the resistors is different.

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Here; R₁ , R₂ and R₃ are connected in series combination.

And this is connected to R₄ in parallel.

The voltage drop in R₁, R₂ & R₄ are 2, 5 and 15 V respectively.

Let V₃ be the voltage drop in R₃

The voltage drop in the series combination is the sum of voltage drops in all the resistors.

So, voltage drop = (2 + 5 + V₃)

We know that in parallel combination, the voltage is same across the resistors.

As the series combination is connected to R₄ in parallel, the voltage drop in the series combination and R₄ is same.

2 + 5 + V₃ = 15

7 + V₃ = 15

V₃ = 15 - 7

V₃ = 8 V

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By Ohm's law : V = IR

Let's apply this formula for the resistor R₃

R₃ = 200 Ω

V₃ = 8 V

8 = I × 200

I = 8/200

I = 1/25 A

∴ The current through R₃ is 1/25 A

We also know that the current flowing through each resistor is same in series connection.

So, current through R₁ = 1/25 A

The voltage drop in R₁ is 2 V

We have to find the resistance.

Again applying Ohm's law,

V = IR

2 = 1/25 × R₁

R₁ = 2 × 25

R₁ = 50 Ω

Therefore, the value of R₁ in the circuit is 50 Ω