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JEE Main 2019 (10 Jan, Shift 2) · Wheatstone Bridge & Metre Bridge

Wheatstone Bridge & Metre Bridge question from JEE Main 2019 (10 Jan, Shift 2)

Q66JEE Main 2019 · 10 JAN · SHIFT 2Medium

Wheatstone Bridge & Metre Bridge

The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80Ω\Omega and 40Ω\Omega, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be -

Figure

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Answer: (a)

  1. Orange, Red, Brown =32×101=320 Ω=R1= 32 \times 10^1 = 320\ \Omega = R_1
  2. Balance: R1R2=R3R4⇒R3=320×4080=160 Ω\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4} \Rightarrow R_3 = \dfrac{320 \times 40}{80} = 160\ \Omega
  3. 160=16×101160 = 16 \times 10^1: Brown (1), Blue (6), Brown (×10\times 10)
  4. So the answer is (a) Brown, Blue, Brown.
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