Question

wow!
I didn't knew that I am so old?!
________________________

Two resistors of resistances 6 Ω and 12 Ω, are connected first in series, and then in parallel in a circuit across a battery of 6 V. Calculate the ratio of the heat produced in the series combination to that of the parallel combination of resistors.

solve this pls​

Answer 1

when connected in series :

R1 = 6 ohm

R2 = 12 ohm

Rs = R1 + R2 = 6 + 12 ohm = 18 ohm

potential difference (V) = 6 V

considering the time for which current flows in the circuit remains same for both combination

Hs = V²t/Rs = (6)²t/18 = 36t/18 J

Hs = 36t/18 J

when connected in parallel :

R1 = 6 ohm

R2 = 12 ohm

1/Rp = 1/6 + 1/12

1/Rp = 3/12

Rp = 4 ohm

V = 6 V

Hp = V²t/Rp = (6)²t/4 = 36t/4 J

Hp = 36t/4 J

Ratio of heat produced in series and heat produced in parallel combination is :

Hs : Hp

36t/ 18 : 36t/ 4

2 : 9

I know the answer is wrong............xD

I just tried randomly for my revision.........xD

you can report this answer and please tell me the correct answer. xD

xD

sahi answer bata do please.............xD

Answer 2

$\huge \red{ \fcolorbox{green}{grey}{required \: answer}}$

resistors :- 6 ohm and 12 ohm

voltage :- 6 V

$\bold { \underline{ \underline \orange{answer}}} \orange→$

2:9

$\bold { \underline{ \underline \orange{solution}}} \orange→$

case (i)

when resistors are connected in series ;

Rs = R1 + R2

→ Rs = 6 + 12

→ Rs = 18 ohm

so, effective resistance in series combination is 18 ohm

we know that..

$\bold{ \color{blue} \: h = \frac{ {v}^{2} t}{r}}$

$\bold{ \implies \: h1 = \frac{ {6}^{2}t }{18} } \\ \\ \bold{ \implies h1 = \frac{36t}{18} } \\ \\ \bold{\implies \: h1 = 2t} \: \: \:$

so heat produced in series combination is 2t joule

case (ii)

when resistors are connected in parallel

we know that;

$\bold{ \frac{1}{Rp} = \frac{1}{r1} + \frac{1}{r2} }$

$\bold { \tt{→ \: \frac{1}{rp} = \frac{1}{6} + \frac{1}{12} }} \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: \frac{1}{Rp} = \frac{3}{12} } \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \implies \: Rp = \frac{12}{3} } \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{\implies \: Rp = 4 \: ohm } \: \: \: \:$

so, effective resistance in parallel combination is 4 ohm

and

$\bold{ \color{blue} \: h = \frac{ {v}^{2} t}{r}}$

$\bold{ \implies \: h2 = \frac{ {6}^{2}t }{4} } \\ \\ \bold{ \implies \: h2 = \frac{36t}{4} } \\ \\ \bold{ \implies \: h2 = 9t} \: \: \:$

so heat produced in parallel combination is 9t joule.

$\bold{ \color{pink}ratio \: of \: h1 \: and \: h2 \: } \\ \\ \bold \blue{ \frac{h1}{h2} = \frac{2t}{9t} } \\ \\ \bold{so \: \: \: h1 : h2 =2 : 9 }$

hence, ratio of heat produced in the series combination to that of the parallel combination of resistors is 2:9