KCET · Physics · Current Electricity
In this circuit, when certain current flows, the heat produced in \(5 \Omega\) is \(4.05 \mathrm{~J}\) in a time t. The heat produced in \(2 \Omega\) coil in the same time interval is
- A \(2.02\)
- B \(5.76\)
- C \(1.44\)
- D \(2.88\)
Answer & Solution
Correct Answer
(D) \(2.88\)
Step-by-step Solution
Detailed explanation
Let, the current distribution is as shown, then by current divider rule,
\(H_{1}=\frac{(6+9)}{(6+9)+5} I=\frac{3}{4} I\)
Let \(\mathrm{H}\) and \(\mathrm{H}_{1}\) be heat produced in \(2 \Omega\) and \(5 \Omega\) resistors respectively in time t, then
\(\mathrm{H}=\mathrm{I}^{2}(2) \mathrm{t}=2 \mathrm{I}^{2} \mathrm{t} \)
\(\text {and } \mathrm{H}_{1}=\mathrm{I}_{1}^{2}(5) \mathrm{t}=\mathrm{II}_{1}^{2} \mathrm{t}=\frac{45}{16} \mathrm{I}^{2} \mathrm{t} \)
\(\therefore \frac{\mathrm{H}}{\mathrm{H}_{\mathrm{I}}}=\frac{32}{45}=\frac{32}{45} \times 4.05 \mathrm{~J}=2.88 \mathrm{~J}\)
\(H_{1}=\frac{(6+9)}{(6+9)+5} I=\frac{3}{4} I\)
Let \(\mathrm{H}\) and \(\mathrm{H}_{1}\) be heat produced in \(2 \Omega\) and \(5 \Omega\) resistors respectively in time t, then
\(\mathrm{H}=\mathrm{I}^{2}(2) \mathrm{t}=2 \mathrm{I}^{2} \mathrm{t} \)
\(\text {and } \mathrm{H}_{1}=\mathrm{I}_{1}^{2}(5) \mathrm{t}=\mathrm{II}_{1}^{2} \mathrm{t}=\frac{45}{16} \mathrm{I}^{2} \mathrm{t} \)
\(\therefore \frac{\mathrm{H}}{\mathrm{H}_{\mathrm{I}}}=\frac{32}{45}=\frac{32}{45} \times 4.05 \mathrm{~J}=2.88 \mathrm{~J}\)
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