Question

Two resistors of 4Ω and 6Ω are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate the following —

i) the power supplied by the battery,

ii) the power used up in each resistor.​

Answer 1

$\huge\fcolorbox{black}{aqua}{Solution:-}$

✏️ Answer :-

★ i) Power supplied by the battery = 15 W.

★ ii) Power dissipated in each resistor, P1 and P2 are 9W and 6W respectively.

✏️ Explanation :-

★ Given :-

• R1 = 4 Ω and,

• R2 = 6 Ω

★ To Find :-

• i) the power supplied by the battery and,

• ii) the power dissipated in each resistor.

★ Rationale :-

∵ Both of the resistors are in parallel,

Thus,

The net equivalent resistance in each resistor will be;

1/Rⁿᵉᵗ = 1/R1 + 1/R2

⇒1/Rⁿᵉᵗ = R2 + R1 /R1R2

⇒Rⁿᵉᵗ = R1R2/ R1 + R2

⇒Rⁿᵉᵗ = 4 × 6 /4 +6

⇒Rⁿᵉᵗ = 2.4 Ω

Now,

Power dissipated,

P = V²/R

⇒P = 36 × 10 /24

⇒P = 15 W (i)

∴ The power supplied by the battery is 15 W .

ii) For R1,

Power dissipated

P1 = V²/R1

⇒P1 = 36/4 = 9W

For R2,

P2 = V²/R² = 6 × 6/6

⇒P2 = 6 W

∴ Power dissipated in each resistor, P1 and P2 are 9W and 6W respectively.

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❏ Commonly made errors :-

• This is often misunderstood by a lot of students. The power formula V²/R has has to be used when the resistors are connected in parallel combination and I²R when connected in series.

❏ Additional Information :-

★ Resistors in Parallel :-

☞ Voltage remains constant but current varies.

☞ I = I1 + I2 + I3....

☞ The net equivalent resistance is given by,

1/Rⁿᵉᵗ = 1/R1 + 1/R2 + 1/R3....

★ Resistors in Series :-

☞ Current remains constant but voltage varies.

☞ V = V1 + V2 + V3....

☞ The total resistance is calculated by,

Rⁿᵉᵗ = R1 + R2 + R3....

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I hope this helps! :)