KCET · Physics · Current Electricity
In the circuit given here, the points A, B and \(C\) are \(70 \mathrm{~V}\), zero, \(10 \mathrm{~V}\) respectively. Then
- A the point \(\mathrm{D}\) will be at a potential of \(60 \mathrm{~V}\)
- B the point \(\mathrm{D}\) will be at a potential of \(20 \mathrm{~V}\)
- C currents in the paths AD, DB and DC are in the ratio of \(1: 2: 3\)
- D currents in the paths \(\mathrm{AB}, \mathrm{DB}\) and \(\mathrm{DC}\) are in the ratio of \(3: 2: 1\)
Answer & Solution
Correct Answer
(C) currents in the paths AD, DB and DC are in the ratio of \(1: 2: 3\)
Step-by-step Solution
Detailed explanation
Applying Kirchhoff's law at point D, we get
\(\mathrm{I}_{1} =\mathrm{I}_{2}+\mathrm{I}_{3} \)
\( \frac{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}}{10} =\frac{\mathrm{V}_{\mathrm{D}}-0}{20}+\frac{\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}}{30} \)
\(\text {or } 70-\mathrm{V}_{\mathrm{D}} =\frac{\mathrm{V}_{\mathrm{D}}+\mathrm{V}_{\mathrm{D}}-10}{2}+\frac{70}{3} \)
\(\Rightarrow \mathrm{V}_{\mathrm{D}} =40 \mathrm{~V} \)
\(\Rightarrow \mathrm{i}_{1} =\frac{70-40}{10}=3 \mathrm{~A}\)
\(\mathrm{i}_{2}=\frac{40-0}{20}=2 \mathrm{~A}\)
and
\(\mathrm{i}_{3}=\frac{40-10}{30}=1 \mathrm{~A}\)
\(\mathrm{I}_{1} =\mathrm{I}_{2}+\mathrm{I}_{3} \)
\( \frac{\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}}{10} =\frac{\mathrm{V}_{\mathrm{D}}-0}{20}+\frac{\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}}{30} \)
\(\text {or } 70-\mathrm{V}_{\mathrm{D}} =\frac{\mathrm{V}_{\mathrm{D}}+\mathrm{V}_{\mathrm{D}}-10}{2}+\frac{70}{3} \)
\(\Rightarrow \mathrm{V}_{\mathrm{D}} =40 \mathrm{~V} \)
\(\Rightarrow \mathrm{i}_{1} =\frac{70-40}{10}=3 \mathrm{~A}\)
\(\mathrm{i}_{2}=\frac{40-0}{20}=2 \mathrm{~A}\)
and
\(\mathrm{i}_{3}=\frac{40-10}{30}=1 \mathrm{~A}\)
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