KCET · Physics · Current Electricity
Three voltmeters A, B and C having resistances \(R, 1.5 R\) and \(3 R\) respectively are used in a circuit as shown. When a potential difference is applied between \(\mathrm{X}\) and \(\mathrm{Y}\), the readings of the voltmeters are \(V_{1}, V_{2}\), and \(V_{3}\) respectively. Then
- A \(\mathrm{V}_{1}=\mathrm{V}_{2}=\mathrm{V}_{3}\)
- B \(\mathrm{V}_{1} < \mathrm{V}_{2}=\mathrm{V}_{3}\)
- C \(\mathrm{V}_{1}>\mathrm{V}_{2}>\mathrm{V}_{3}\)
- D \(V_{1}>V_{2}=V_{3}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{V}_{1}=\mathrm{V}_{2}=\mathrm{V}_{3}\)
Step-by-step Solution
Detailed explanation
\(\text {Here, } \mathrm{V}_{2}=\mathrm{V}_{3} \)
\(\text {ie, } \mathrm{i}_{2} \times 1.5 \mathrm{R}=3 \mathrm{R} \times \mathrm{i}_{3} \)
\(\text {and } \mathrm{i}_{2}+\mathrm{i}_{3}=\mathrm{i} \)
\(\Rightarrow \mathrm{i}_{2}=\frac{2 \mathrm{i}}{3} \text { and } \mathrm{i}_{3}=\frac{\mathrm{i}}{3} \)
\(\text {Now, } \mathrm{V}_{1}=\mathrm{IR} \)
\(\mathrm{V}_{2}=\frac{2 \mathrm{i}}{3} \times 1.5 \mathrm{R}=\mathrm{iR} \)
\(\text {ie, } \mathrm{V}_{3}=\frac{\mathrm{i}}{3} \times 3 \mathrm{R}=\mathrm{iR}\)
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