Question

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 $R_2\text{and}{R}_4$ are $80\Omega \text{and}40\Omega$ 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

• A. Brown, Blue, Brown
• B. Brown, Blue, Black
• C. Red, Green, Brown
• D. Grey, Black, Brown

Answer 1

The correct option is A Brown, Blue, Brown

Given colour code of resistance,
$R_1=$Orange,Red and Brown
$\therefore R_1=32\times 10=320\Omega$
Using balanced wheatstone bridge principle,
$\Rightarrow \frac{R_1}{R_3}=\frac{R_2}{R_4}\Rightarrow \frac{320}{R_3}=\frac{80}{40}$
$\therefore R_3=160$
i.e. colour code for $R_3$ is Brown, Blue and Brown.