Question

Two identical resistor each of resistance 12ohm are connected A. in series B. in parallel in turn to a battery of 6V. calculate the ratio of power consumed in the combination of resistor in two cases. A. 1;4 b. 4;1c. 2;7 d. 1;6​

Answer 1

Answer :-

Ratio of power consumed in the combination of resistor in two cases is 1 : 4 [Option.A] .

Explanation :-

For series connection :-

Equivalent resistance (Rₛ) :-

⇒ Rₛ = R₁ + R₂

⇒ Rₛ = (12 + 12) Ω

⇒ Rₛ = 24 Ω

Power consumed :-

⇒ Pₛ = V²/Rₛ

⇒ Pₛ = (6)²/24

⇒ Pₛ = 36/24

⇒ Pₛ = 1.5 W

For parallel connection :-

Equivalent resistance (Rₚ) :-

⇒ 1/Rₚ = 1/R₁ + 1/R₂

⇒ 1/Rₚ = 1/12 + 1/12

⇒ 1/Rₚ = 1/6

⇒ Rₚ = 6 Ω

Power consumed :-

⇒ Pₚ = V²/Rₚ

⇒ Pₚ = (6)²/6

⇒ Pₚ = 36/6

⇒ Pₚ = 6 W

Ratio of power consumed :-

= Rₛ : Rₚ

= 1.5 : 6

= 1.5/6

= 1/4

= 1 : 4

Answer 2

Given:-

• Two identical resistor each of resistance 12ohm are connected A in series B. In parallel in turn to a battery of 6V.

To Find:-

• Ratio of power consumed in the combination of resistor in two cases.

Formula Used:-

• ${\boxed{\sf{R_{(s)}=R_1+R_2}}}$

• ${\boxed{\sf{\dfrac{1}{R_{(p)}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}}}$

• ${\boxed{\sf{P=\dfrac{V^2}{R}}}}$

Here,

• $\bf R_{(s)}=$ Equivalent Resistance in series.

• $\bf R_{(p)}=$ Equivalent Resistance in parallel.

• $\bf P=$ Power Consumed

• $\bf V=$ Potential Difference

• $\bf R=$ Resistance

Solution:-

A) Series connection:-

Using Formula,

$\sf :\implies\:R_{(s)}=R_1+R_2$

$\sf :\implies\:R_{(s)}=12+12$

$\sf :\implies\:R_{(s)}=24\:ohm$

Now,

$\sf :\implies\:P_{(s)}=\dfrac{V^2}{R_{(s)}}$

$\sf :\implies\:P=\dfrac{6^2}{24}$

$\sf :\implies\:P=\dfrac{36}{24}$

$\sf :\implies\:P=\dfrac{3}{2}\:W$

B) Parallel connection:-

$\sf :\implies\:\dfrac{1}{R_{(p)}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\sf :\implies\:\dfrac{1}{R_{(p)}}=\dfrac{1}{12}+\dfrac{1}{12}$

$\sf :\implies\:\dfrac{1}{R_{(p)}}=\dfrac{1+1}{12}$

$\sf :\implies\:\dfrac{1}{R_{(p)}}=\dfrac{2}{12}$

$\sf :\implies\:\dfrac{1}{R_{(p)}}=\dfrac{1}{6}$

$\sf :\implies\:R_{(p)}=6\:ohm$

Now,

$\sf :\implies\:P_{(p)}=\dfrac{V^2}{R_{(p)}}$

$\sf :\implies\:P=\dfrac{6^2}{6}$

$\sf :\implies\:P=\dfrac{36}{6}$

$\sf :\implies\:P=6\:W$

Ratio of power consumed:-

$\sf :\implies\:R_{(s)} : R_{(p)}=\dfrac{3}{2} : 6$

$\sf :\implies\:\dfrac{R_{(s)}}{R_{(p)}}=\dfrac{\dfrac{3}{2}}{6}$

$\sf :\implies\:\dfrac{R_{(s)}}{R_{(p)}}=\dfrac{3}{2\times6}$

$\sf :\implies\:\dfrac{R_{(s)}}{R_{(p)}}=\dfrac{1}{2\times2}$

$\sf :\implies\:\dfrac{R_{(s)}}{R_{(p)}}=\dfrac{1}{4}$

Hence, The Ratio of power consumed in the combination of resistor in two cases is 1 : 4.

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