JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 3)
The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80$$\Omega $$ 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 R3, would be -
_10th_January_Evening_Slot_en_3_1.png)
Brown, Blue, Brown
Grey, Black, Brown
Red, Green, Brown
Brown, Blue, Black
Explanation
R1 = 32 $$ \times $$ 10 = 320
for wheatstone bridge
$$ \Rightarrow $$ $${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$$
$${{320} \over {{R_3}}} = {{80} \over {40}}$$
$${R_3} = 160$$
$$ \therefore $$ Correct answer is Brown Blue Brown
for wheatstone bridge
$$ \Rightarrow $$ $${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$$
$${{320} \over {{R_3}}} = {{80} \over {40}}$$
$${R_3} = 160$$
$$ \therefore $$ Correct answer is Brown Blue Brown
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