The Wheatstone bridge shown in the figure below, gets balanced when the carbon resistor used as R 1 has the…

The Wheatstone bridge shown in the figure below, gets balanced when the carbon resistor used as R1 has the colour code (orange, red, brown). The resistors R2 and R4 are 80 Ω and 40 Ω, 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

  1. brown, blue, black.
  2. brown, blue, brown.
  3. grey, black, brown.
  4. red, green, brown.

Solution

Colour band A and B indicate the first two significant figures of resistance in 'ohm', while the C band gives the decimal multiplier i.e., the number of zeros that follows the two significant figures A and B.

The last band D indicates the tolerance in percentage about the indicated value or in other word, it represents the percentage accuracy of the indicated value. 

For R1, the given colour of the rings of a carbon resistor are orange, red, brown then, the value of resistance is

R1=32×101

Condition for balance in the Wheat stone's bridge is

R1R3=R2R4

So, 320R3=8040

R3=160 Ω, so colour coding of resistor R3 is

Brown1

Blue6

Brown1
16×101160 Ω

Asked in: JEE Main 2019 (10 Jan Shift 2)

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