The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R1 has the colour code ( Orange, Red, Brown). The resistors R2 and R4 are 80Ω and 40 Ω 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 :

Question

The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R1 has the colour code ( Orange, Red, Brown). The resistors R2 and R4 are 80Ω and 40 Ω 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 : (A) Red, Green, Brown (B) Brown, Blue, Brown (C) Grey, Black, Brown (D) Brown, Blue, Black

Tardigrade Admin · Accepted Answer

R1 = 32 ×10 = 320 for wheat stone bridge ⇒ (R1/R3) = (R2/R4) (320/R3) = (80/40)

Red, Green, Brown

Brown, Blue, Brown

Grey, Black, Brown

Brown, Blue, Black

R1​=32×10=320 for wheat stone bridge ⇒R3​R1​​=R4​R2​​ R3​320​=4080​

$R_{1} = 32 \times 10 = 320$
for wheat stone bridge
$\Rightarrow \frac{R _{1}}{R _{3}} = \frac{R _{2}}{R _{4}}$
$\frac{320}{R _{3}} = \frac{80}{40}$