KCET · Physics · Current Electricity
In the circuit diagram, heat produces in \(R, 2 R\) and \(1.5 R\) are in the ratio of
- A \(4: 2: 3\)
- B \(8: 4: 27\)
- C \(2: 4: 3\)
- D \(27: 8: 4\)
Answer & Solution
Correct Answer
(B) \(8: 4: 27\)
Step-by-step Solution
Detailed explanation
We have, \(l_{1}=\frac{I \times 2 R}{3 R}=\frac{2 I}{3}\)
\(\therefore H_{1}=l_{1}^{2} R=\frac{4 l^{2}}{9} \times R...(i) \)
\( \text {Also, } I_{2}=\frac{I \times R}{3 R}=\frac{l}{3} \)
\( \therefore H_{2}=I_{2}^{2}(2 R)=\frac{l^{2}}{9} \times 2 R...(ii) \)
\( \text {and } H_{3}=I^{2}(1.5 R)...(iii)\)
From Eqs. (i), (ii) and (iii),
\(H_{1}: H_{2}: H_{3} =\frac{4 l^{2}}{9} \times R: \frac{I^{2}}{9} \times 2 R: I^{2} \times 1.5 R \)
\( =\frac{4}{9}: \frac{2}{9}: 1.5=4: 2: 13.5 \)
\( =8: 4: 27\)
\(\therefore H_{1}=l_{1}^{2} R=\frac{4 l^{2}}{9} \times R...(i) \)
\( \text {Also, } I_{2}=\frac{I \times R}{3 R}=\frac{l}{3} \)
\( \therefore H_{2}=I_{2}^{2}(2 R)=\frac{l^{2}}{9} \times 2 R...(ii) \)
\( \text {and } H_{3}=I^{2}(1.5 R)...(iii)\)
From Eqs. (i), (ii) and (iii),
\(H_{1}: H_{2}: H_{3} =\frac{4 l^{2}}{9} \times R: \frac{I^{2}}{9} \times 2 R: I^{2} \times 1.5 R \)
\( =\frac{4}{9}: \frac{2}{9}: 1.5=4: 2: 13.5 \)
\( =8: 4: 27\)
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