The Wheatstone bridge shown in Fig. here, 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:
Correct Option: 1
(1) Given, colour code of resistance,
$\mathrm{R}_{1}=$ Orange, Red and Brown
$\therefore \mathrm{R}_{1}=32 \times 10=320$
using balanced wheatstone bridge principle,
$\frac{\mathrm{R}_{1}}{\mathrm{R}_{3}}=\frac{\mathrm{R}}{\mathrm{R}_{4}} \Rightarrow \frac{320}{\mathrm{R}_{3}}=\frac{80}{40}$
$\therefore \mathrm{R}_{3}=160$ i.e. colour code for $\mathrm{R}_{3}$ Brown, Blue and Brown
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