Question

calculate
(i) value of current flowing in each resistor,
(ii) the total current in the circuit
(iii) the total effective resistance of the circuit
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Answer 1

Explanation:

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Answer 2

Explanation:

The three resistors are connected in parallel as the current flowing through each of them is not equal.

Let

• R₁ = 5 Ω
• R₂ = 30 Ω
• R₃ = 10 Ω

In a parallel combination, the effective resistance (R) is given by

  $\boxed{\longrightarrow \tt \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}$

Substitute the values,

$\rm \dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{30}+\dfrac{1}{10} \\\\ \rm \dfrac{1}{R}=\dfrac{6}{30}+\dfrac{1}{30}+\dfrac{3}{30} \\\\ \rm \dfrac{1}{R}=\dfrac{6+1+3}{30} \\\\ \rm \dfrac{1}{R}=\dfrac{10}{30} \\\\ \boxed{\longrightarrow \rm R=3 \Omega}$

Therefore, the total effective resistance of circuit is 3 Ω

Given, voltage across the circuit is 6 V

From Ohm's law, V = IR

So, 6 = I × 3

 I = 6/3

 I = 2 A

∴ The total current in the circuit is 2 A

• In parallel combination of resistors, the potential difference across each resistor is constant.

i.e., V₁ = V₂ = V₃ = V = 6 V

Current through 5 Ω resistor :

V₁ = I₁R₁

6 = I₁ × 5

I₁ = 6/5

I₁ = 1.2 A

Current through 30 Ω resistor :

V₂ = I₂R₂

6 = I₂ × 30

I₂ = 6/30

I₂ = 1/5

I₂ = 0.2 A

Current through 10 Ω resistor :

V₃ = I₃R₃

6 = I₃ × 10

I₃ = 6/10

I₃ = 0.6 A