Question

In an electric circuit , two resistors of 2 ohm and 4 ohm are connected in parallel to 6v battery . Calculate the heat dissipated by the 4 ohm resistor in 5 sec

Answer 1

Answer :

Two identical resistors of resistances 2Ω and 4Ω are connected in parallel to a battery of 4V.

We have to find heat dissipated by the 4Ω resistor in 5s$.$

Concept : Potential difference across each resistor in parallel connection remains same.

Formula of heat dissipated by resistor connected in parallel is given by

• H = V²/R × t

Let's calculate ☃

➝ H = V²/R × t

➝ H = 6²/4 × 5

➝ H = 9×5

➝ H = 45J

Second method [Follow this steps]

• Find current flow in 4Ω resistor using ohm's law.
• Apply H = I²R × t ! You'll get the same answer :D

Answer 2

$\huge\sf\pink{Answer}$

☞ Heat Dispersed is 45 J

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$\huge\sf\blue{Given}$

✭ Two resistors of 2Ω & 4Ω are connected in parallel

✭ Voltage = 6 V

✭ Time = 5 sec

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$\huge\sf\gray{To \:Find}$

◈ The heat dispersed?

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$\huge\sf\purple{Steps}$

So here we shall first find the current through the circuit and for that we shall use the Ohm's law,that is,

$\underline{\boxed{\sf V = IR}}$

Substituting the given values,

➝ $\sf V = IR$

➝ $\sf 6 = I\times 4$

➝ $\sf \dfrac{6}{4} = I$

➝ $\sf \red{I = 1.5 \ A}$

So now we shall find the heat dispersed with the help of,

$\underline{\boxed{\sf H = I^2 Rt}}$

Substituting the values,

➳ $\sf H = 1.5^2 \times 4 \times 5$

➳ $\sf H = 2.25\times 4 \times 5$

➳ $\sf \orange{H = 45 \ J}$

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