Question

The figure shown below is a balanced Wheetstone bridge. The resistor used as $R_1$, has the colour code ( orange, red , brown). The resistors $R_2$ and $R_4$ are of $80\Omega and40\Omega$ respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code of the resistors $R_3$ will be

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

Answer 1

The correct option is B Brown, Blue, Brown

The value of $R_1$ ( Orange, Red, Brown)

$R_1=32\times 10=320\Omega$

For a balanced Wheatstone bridge, $\frac{R_1}{R_2}=\frac{R_3}{R_4}$

$\Rightarrow \frac{320}{80}=\frac{R_3}{40}$

$\therefore R_3=160\Omega$

Colour coding is



Hence, option (b) is the correct answer.