JEE PYQ: Current Electricity Question 65
Question 65 - 2019 (10 Jan 2019 Shift 2)
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 $R_2$ and $R_4$ 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 $R_3$, would be:
(1) Brown, Blue, Brown
(2) Brown, Blue, Black
(3) Red, Green, Brown
(4) Grey, Black, Brown
Show Answer
Answer: (1)
Solution
Orange, Red, Brown: $R_1 = 32 \times 10 = 320,\Omega$. For balanced bridge: $\frac{R_1}{R_2} = \frac{R_3}{R_4}$, $\frac{320}{80} = \frac{R_3}{40}$. $R_3 = 160,\Omega$. Brown(1), Blue(6), Brown($\times 10$): $= 160,\Omega$.
BETA
