Question

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.


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Answer 1

$\huge \bold\color{green}꧁{ \overbrace{ \underbrace \mathfrak \red{answer}}}꧂$

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

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