Question

Two resistors each of 10 Ω are connected in i) series ii) and then in parallel to a battery of 6 V. Calculate the ratio of power consumed in the combination of resistor in two case.​

Answer 1

Answer :-

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

Explanation :-

For series connection :-

Equivalent resistance (R) :-

⇒ R = R₁ + R₂

⇒ R = (10 + 10) Ω

⇒ R = 20 Ω

Power consumed (P₁) :-

⇒ P₁ = V²/R

⇒ P₁ = (6)²/20

⇒ P₁ = 36/20

⇒ P₁ = 1.8 W

For parallel connection :-

Equivalent resistance (R) :-

⇒ 1/R = 1/R₁ + 1/R₂

⇒ 1/R = 1/10 + 1/10

⇒ 1/R = 2/10

⇒ R = 5 Ω

Power consumed (P₂) :-

⇒ P₂ = V²/R

⇒ P₂ = (6)²/5

⇒ P₂ = 36/5

⇒ P₂ = 7.2 W

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Now, ratio of power consumed :-

= P₁ : P₂

= 1.8 : 7.2

= 1.8/7.2

= 1/4

= 1 : 4

Answer 2

Given:-

• Two resistors = 10 Ω each

• Potential Difference = 6V

• Resistors connected (i) in series and (ii) in parallel

To Find:-

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

Formula used:-

• $R_{(series)}=R_1+R_2$

• $\dfrac{1}{R_{(parallel)}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

• $I=\dfrac{V}{R}$

• $Power\:=VI$

where,

• $R_{(series)}$= Equivalent resistance in series

• $R_{(parallel)}$= Equivalent resistance in parallel

• $R_1\:and\:R_2$= Given resistance

• I = Current

• V = Potential Difference

Solution:-

Firstly,

(i) In series

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

$\implies\:R_{(series)}=10Ω+10Ω$

${\boxed{\implies\:R_{(series)}=20Ω}}$

Now,

$\implies\:I_{(series)}=\dfrac{V}{R_{series}}$

$\implies\:I_{(series)}=\dfrac{6}{20}$

$\implies\:I_{(series)}=\dfrac{3}{10}$

${\boxed{\implies\:I_{(series)}=0.3\:A}}$

Now,

(ii) In parallel

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

$\implies\:\dfrac{1}{R_{(parallel)}}=\dfrac{1}{10}+\dfrac{1}{10}$

$\implies\:\dfrac{1}{R_{(parallel)}}=\dfrac{1+1}{10}$

$\implies\:\dfrac{1}{R_{(parallel)}}=\dfrac{2}{10}$

$\implies\:\dfrac{1}{R_{(parallel)}}=\dfrac{1}{5}$

${\boxed{\implies\:R_{(parallel)}=5Ω}}$

Now,

$\implies\:I_{(parallel)}=\dfrac{V}{R_{(parallel)}}$

$\implies\:I_{(parallel)}=\dfrac{6}{5}$

${\boxed{\implies\:I_{(parallel)}=1.2\:A}}$

Finally,

Ratio of power = $\dfrac{P_{(series)}}{P_{(parallel)}}$

Ratio of power = $\dfrac{V×I_{(series)}}{V \times I_{(parallel)}}$

Ratio of power = $\dfrac{6 \times 0.3}{6×1.2}$

Ratio of power = $\dfrac{1.8}{7.2}$

Ratio of power = $\dfrac{18}{72}$

Ratio of power = $\dfrac{1}{4}$

Hence, The Ratio of the power of these two cases is 1:4.

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