Question

The four resistors 20ohm,40ohm,20+x ohm,80ohm if the bridge is balanced respectively form a Wheatstone bridge. Find the value of x

Answer 1

The value of x = 20 ohm.

• According to the principle of  wheatstone bridge,

(Resistor r1) / (Resistor r2) = (Resistor r3) / (Resistor r4)

• Given ,

Value of Resistor r1 = 20 ohm

Value of Resistor r2 = 40 ohm

Value of Resistor r3 = (20+x) ohm

Value of Resistor r4 = 80 ohm

• Then by using principal of wheatstone bridge,

(20 / 40) =  [(20 + x)/80]

½ =  [(20 + x)/80]

40 = 20 + x

x = 20 ohm

Answer 2

The value of x for the Wheatstone bridge condition is 140 ohm .

Explanation:

Given as :

The four Resistor are in balance condition in a form of Wheatstone bridge .

The measure of resistor  A = $R_1$ = 20 ohm

The measure of resistor  B = $R_2$ = 40 ohm

The measure of resistor  C = $R_3$ = ( 20 + x ) ohm

The measure of resistor  D = $R_4$ = 80 ohm

According to question

Wheatstone bridge - A simple circuit for measuring an unknown resistance by connecting it so as to form a quadrilateral with three known resistances and applying a voltage between a pair of opposite corners.

Wheatstone bridge at balance condition

From figure

resistance across AB × resistance across CD  = resistance across BC × resistance across DA

i.e   $R_1$  ×  $R_3$   =  $R_2$  ×  $R_4$

Or, ( 20 $\Omega$ ) × (  20 + x ) $\Omega$  = ( 40 $\Omega$ ) × ( 80 $\Omega$ ) ×

Or,  400 + 20 x  = 3200

Or,             20 x = 3200 - 400

Or,             20 x = 2800

∴                     x = $\dfrac{2800}{20}$

i.e                   x = 140  $\Omega$

So, The value of x =  140  $\Omega$

Hence, The value of x for the Wheatstone bridge condition is 140 ohm . Answer