Question

Example 11. Two 40 ohm resistors and a 20 ohm resistor are all connected in parallel with a 12 V power supply. Calculate their effective resistance and the current through each resistor. What is the current flowing through the supply?​

Answer 1

Answer :-

→ Effective resistance = 10 Ω

→ Current through 40 Ω resistors = 0.3 A

→ Current through 20 Ω resistor = 0.6 A

→ Current through the supply = 1.2 A

Explanation :-

We have :-

• 1st Resistance (R₁) = 40 Ω

• 2nd Resistance (R₂) = 40 Ω

• 3rd Resistance (R₃) = 20 Ω

• Potential difference (V) = 12 V

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For parallel connection, we have effective Resistance as :-

1/R = 1/R₁ + 1/R₂ + 1/R₃

⇒ 1/R = 1/40 + 1/40 + 1/20

⇒ 1/R = (1 + 1 + 2)/40

⇒ 1/R = 4/40

⇒ 4R = 40

⇒ R = 40/4

⇒ R = 10 Ω

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Now, let's calculate the current through each resistor .

I₁ = V/R₁

⇒ I₁ = 12/40

⇒ I₁ = 0.3 A

I₂ = V/R₂

⇒ I₂ = 12/40

⇒ I₂ = 0.3 A

I₃ = V/R₃

⇒ I₃ = 12/20

⇒ I₃ = 0.6 A

Finally, total current flowing through the supply will be :-

I = I₁ + I₂ + I₃

⇒ I = (0.3 + 0.3 + 0.6) A

⇒ I = (0.6 + 0.6) A

⇒ I = 1.2 A

Answer 2

Question:

• Two 40 Ω resistors and a 20 Ω resistor are all connected in parallel with a 12 V power supply. Calculate their effective resistance and the current through each resistor. What is the current flowing through the supply?

Answer:

• Effective resistance = 10 Ω
• Current through 40 Ω resistors = 0.3 A
• Current through 20 Ω resistors = 0.6 A
• Current flowing through the supply = 1.2 A

Explanation:

Given that:

• We have three resistors.
• Two 40 Ω resistors.
• One 20 Ω resistor.
• All are connected in parallel with 12 V supply.

To Find:

• Effective resistance?
• Current through each resistor?
• Current flowing through the supply?

Solution:

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❶ Finding effective resistance :-

We know that,

• In parallel connection, it is given by ::

⋆ $\boxed{\bf{\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}}}}$

Putting values in formula we get,

➻ $\sf \dfrac{1}{R_{eq}} = \dfrac{1}{40} + \dfrac{1}{40} + \dfrac{1}{20}$

➻ $\sf \dfrac{1}{R_{eq}} = \dfrac{1 + 1 + 2}{40}$

➻ $\sf \dfrac{1}{R_{eq}} = {\cancel{\dfrac{4}{40}}}$

➻ $\sf \dfrac{1}{R_{eq}} = \dfrac{1}{10}$

By doing cross multiplication we get,

➻ $\underline{\boxed{\bf{R_{eq} = 10\:\Omega}}}$

• Therefore, effective resistance is 10 Ω.

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❷ Finding current through each resistor :-

We know that,

• Ohm's law ::

⋆ $\boxed{\bf{V = IR}}$

Putting values in formula we get,

➜ $\sf 12 = I_{1} \:\times\: 40$

➜ $\sf I_{1} = {\cancel{\dfrac{12}{40}}}$

➜ $\underline{\boxed{\bf{I_{1} = 0.3\:A}}}$

And,

➜ $\sf 12 = I_{2} \:\times\: 40$

➜ $\sf I_{2} = {\cancel{\dfrac{12}{40}}}$

➜ $\underline{\boxed{\bf{I_{2} = 0.3\:A}}}$

Also,

➜ $\sf 12 = I_{3} \:\times\: 20$

➜ $\sf I_{3} = {\cancel{\dfrac{12}{20}}}$

➜ $\underline{\boxed{\bf{I_{3} = 0.6\:A}}}$

• Therefore, current through 40 Ω resistors is 0.3 A and current through 20 Ω resistors is 0.6 A.

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❸ Finding current flowing through the supply :-

We know that,

⋆ $\boxed{\bf{Total\:current = I_{1} + I_{2} + I_{3}}}$

Putting values in equation we get,

⇰ $\sf Total\:current = 0.3 + 0.3 + 0.6$

⇰ $\underline{\boxed{\bf{Total\:current = 1.2\:A}}}$

• Therefore, current flowing through the supply is 1.2 A.

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