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NEET 2005 (AIPMT) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from NEET 2005 (AIPMT)

Q197NEET 2005 · AIPMTMedium

Kirchhoff's Laws & Measuring Instruments

For the network shown in the figure the value of the current ii is

Figure

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Answer: (d)

  1. Ratios 42=63\frac{4}{2} = \frac{6}{3}, so the bridge is balanced and the middle 4 Ω4\ \Omega carries no current.
  2. Paths: 4+2=6 Ω4 + 2 = 6\ \Omega and 6+3=9 Ω6 + 3 = 9\ \Omega in parallel.
  3. R=6×915=185 ΩR = \frac{6 \times 9}{15} = \frac{18}{5}\ \Omega
  4. i=VR=5V18i = \frac{V}{R} = \frac{5V}{18}
  5. So the answer is (d).
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