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JEE Main 2020 (3 Sep, Shift 2) · Kirchhoff's Laws & Measuring Instruments

Kirchhoff's Laws & Measuring Instruments question from JEE Main 2020 (3 Sep, Shift 2)

Q147JEE Main 2020 · 3 SEP · SHIFT 2Medium

Kirchhoff's Laws & Measuring Instruments

Two resistors 400Ω\Omega and 800Ω\Omega are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 kΩ\Omega across 400 Ω\Omega resistor is close to:

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Answer: (b)

  1. 400 Ω∥10 kΩ=400×1000010400≈384.6 Ω400\ \Omega \| 10\ \text{k}\Omega = \dfrac{400 \times 10000}{10400} \approx 384.6\ \Omega
  2. Total R=384.6+800=1184.6 ΩR = 384.6 + 800 = 1184.6\ \Omega
  3. V=6×384.61184.6≈1.95 VV = 6 \times \dfrac{384.6}{1184.6} \approx 1.95\text{ V}
  4. So the answer is (b) 1.95 V.
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