Q 12-03-248JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
The Wheatstone bridge shown in the figure below, gets balanced when the carbon resistor used as $R_1$ has the colour code (orange, red, brown). The resistors $R_2$ and $R_4$ are $80\ \Omega$ and $40\ \Omega$, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as $R_3$, would be
Answer: (B) brown, blue, brown
Orange $= 3$, red $= 2$, brown $= \times10$: $R_1 = 320\ \Omega$.
At balance $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$:
$$R_3 = \frac{320\times40}{80} = 160\ \Omega = 16\times10^1\ \Omega$$
Colour code: brown $(1)$, blue $(6)$, brown $(\times10)$.
Solution by Sreeraj P, M.Sc Physics