question_answer

The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as \[{{R}_{1}}\] has the colour code (Orange, Red, Brown).  he resistors \[{{R}_{2}}\text{ }and\text{ }{{R}_{4}}\text{ }are\text{ }80\Omega \,\,and\text{ }40\Omega \] , respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as Rs, would be- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

A) Brown, Blue, Brown                   

B) Grey, Black, Brown

C) Red, Green, Brown                     

D) Brown, Blue, Black

Correct Answer: B

Solution :

Color code for \[{{R}_{1}}\] orange, red, brown \[\therefore \,\,\,\text{ }{{R}_{1}}=32\times 10=320\text{ }\Omega \] Balanced condition \[\frac{{{R}_{1}}}{{{R}_{3}}}\,\,=\,\,\frac{{{R}_{2}}}{{{R}_{4}}}\] \[\frac{320}{{{R}_{3}}}\,\,=\,\,\frac{80}{40}\] \[~{{R}_{3}}=160\text{ }\Omega \] Corresponding color code for \[{{R}_{3}}\] = Brown, Blue and Brown