Electricity immediately gives us the relations I = I and I = I . Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first loop gives – I R + + I R = ( I g = ) ( . ) and the second loop gives, upon using I = I , I = I I R + – I R = ( .
) From Eq. ( . ), we obtain, I I whereas from Eq. ( .
), we obtain, I I Hence, we obtain the condition [ . (a)] This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection. The Wheatstone bridge and its balance condition provide a practical method for determination of an unknown resistance. Let us suppose we have an unknown resistance, which we insert in the fourth arm; R is thus not known.
Keeping known resistances R and R in the first and second arm of the bridge, we go on varying R till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance R is given by, R R [ . (b)] A practical device using this principle is called the meter bridge . It will be discussed in the next section.
Example . The four arms of a Wheatstone bridge (Fig. . ) have the following resistances: AB = Ω , BC = Ω , CD = Ω, and DA = Ω .
FIGURE . FIGURE . E XAMPLE . E XAMPLE .
A galvanometer of Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of V is maintained across AC. Solution Considering the mesh BADB, we have I + I g – I = or I + I g – I = [ . (a)] Considering the mesh BCDB, we have ( I – I g ) – I g – ( I + I g ) = I – I g – I = I – I g – I = [ .
(b)] Considering the mesh ADCEA, I + ( I + I g ) = I + I g = I + I g = [ . (c)] Multiplying Eq. ( .84b) by I – I g – I = [ . (d)] From Eqs.
( .84d) and ( .84a) we have I g – I = I = . I g [ . (e)] Substituting the value of I into Eq. [ .
(c)], we get ( . I g ) + I g = . I g = I g = . mA.