Q. 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 :

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Solution:

$R_{1} = 32 \times10 = 320$
for wheat stone bridge
$ \Rightarrow \frac{R_{1}}{R_{3}} = \frac{R_{2}}{R_{4}}$
$\frac{320}{R_{3}} = \frac{80}{40} $

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