| The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R, has the colour code (Orange, Red, Brown), The resistors \[{{R}_{2}}\] and \[{{R}_{4}}\] are \[80\Omega \] and\[40\Omega ,\], respectively. |
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| 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
Correct Answer: A
Solution :
| \[{{R}_{1}}=32\times 10=320\Omega \] |
| \[{{R}_{3}}=\frac{{{R}_{y}}}{{{R}_{2}}}\times {{R}_{1}}=\frac{40\times 320}{80}=160\Omega \] |
| \[\therefore \]Colour code or \[{{R}_{3}}\] be Brown, Blue, Brown. |
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