Question

13.Obseve the following circuit and calculate
1. Effective resistance
2. Calculate current drawn from cell.




anyone who can solve this ​

Answer 1

Answer:

⇒ The effective resistance of the circuit is 3 Ω

⇒ The current drawn from the cell is 0.5 A

Explanation:

To know :  

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in series, the equivalent resistance (R) is given by

      R = R₁ + R₂ + R₃ + ...  

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in parallel, the equivalent resistance (R) is given by

      1/R = 1/R₁ + 1/R₂ + 1/R₃ + ...

Solution :

Let

• R₁ = 1 Ω
• R₂ = R₃ = 2 Ω
• R₄ = 1 Ω

The two resistors of resistance 2 Ω each i.e., R₂ and R₃ are connected in parallel.

Let R₂₃ be its effective resistance.

$\sf \dfrac{1}{R_{23}}=\dfrac{1}{R_2}+\dfrac{1}{R_3} \\\\ \sf \dfrac{1}{R_{23}}=\dfrac{1}{2}+\dfrac{1}{2} \\\\ \sf \dfrac{1}{R_{23}}=\dfrac{1+1}{2} \\\\ \sf \dfrac{1}{R_{23}}=\dfrac{2}{2} \\\\ \sf R_{23}=1 \Omega$

So, the effective resistors of the resistors connected in parallel is 1 Ω

Now, R₁ , R₂₃ and R₄ are connected in series.

Let R₁₂₃₄ be the effective resistance of the circuit.

R₁₂₃₄ = R₁ + R₂₃ + R₄

R₁₂₃₄ = 1 + 1 + 1

R₁₂₃₄ = 3 Ω

The effective resistance of the circuit is 3 Ω

Given, the voltage across the cell is 1.5 V

From Ohm's law, V = IR

where

I is the current drawn from the cell/flowing through the circuit

R is the effective resistance of the circuit

Substituting,

1.5 = I × 3

I = 1.5/3

I = 0.5 A

The current drawn from the cell is 0.5 A