In the following diagram, on what value shall we keep R₂ so that the current across galvanometer V_G is zero?

Concept:

• Wheatstone Bridge: In the below diagram, BD is Wheatstone Bridge when there is no current across it. This happens when

\(\frac{R_1}{R_2} = \frac{R_3}{R_4}\)

• The above figure is an example of Wheatstone Bridge where the Galvanometer is acting as the bridge. 
• The zero deflection of the galvanometer forms Wheatstone Bridge when no current is in Galvanometer.
• This is used to find unknown resistances. 

Calculation:

So, the given condition for Wheatstone is satisfied when 

\(\frac{R_1}{R_2} = \frac{R_3}{R_4}\)

\(\implies R_2 = \frac{R_1R_4}{R_3}\)

So, the correct option is \(R_2 = \frac{R_1R_4}{R_3}\)

Additional Information

• Meter Bridge: It is an electrical instrument used to measure an unknown resistance.
• It works on two principles, one is Wheatstone Bridge, and the other is 'direct proportionality relationship between length and resistance of a wire.'

• Working of meter Bridge: AC is long resistance wire 100 cm long.. Varying the position of tapping point B, the bridge is balanced.
• Bridge Balanced means the galvanometer is showing zero resistance and the condition satisfying wheatstone Bridge.
• If B is the balanced position of bridge AB = l, BC (100 – l) so that

\(\frac{Q}{P} = \frac{{\left( {100\; - \;l} \right)}}{l}\)