The Wheatstone bridge shown in the figure get | vedclass

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(A) For a balanced Wheatstone bridge,the condition is $\frac{R_1}{R_2} = \frac{R_3}{R_4}$.The colour code for $R_1$ is (Orange,Red,Brown). Using the s

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Brown,Blue,Brown

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(A) For a balanced Wheatstone bridge,the condition is $\frac{R_1}{R_2} = \frac{R_3}{R_4}$.The colour code for $R_1$ is (Orange,Red,Brown). Using the standard colour code table: Orange = $3$,Red = $2$,Brown = $10^1$. Thus,$R_1 = 32 	imes 10^1 = 320 \, \Omega$.Given $R_2 = 80 \, \Omega$ and $R_4 = 40 \, \Omega$.From the balanced condition,$R_3 = R_1 	imes \frac{R_4}{R_2} = 320 	imes \frac{40}{80} = 320 	imes 0.5 = 160 \, \Omega$.For $R_3 = 160 \, \Omega$,the colour code is: $1$ (Brown),$6$ (Blue),$10^1$ (Brown). Therefore,the colour code is (Brown,Blue,Brown).

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