Application of Kirchhoff laws to Wheatstone Network

Four resistance P, Q, S and R from closed network ABCD. A cell of emf E is connected between A and C. A galvanometer of resistance G is connected between B and D.

Let I be the current along EA, I₁ along AB and Ig along BD. By Kirchhoff’s first law, the current along the various branches will be as shown.

We can obtain an expression for the current through the galvanometer ig by applying Kirchhoff’s voltage law (KVL).

Applying KVL to the closed mesh ABDA,
I₁P + I_g G – (I – I₁)R = 0 or (P +R)I₁ + G I_g = IR_______________________________… (1)

Applying KVL to the closed mesh BCDB,
(I₁ – I_g)Q – (I – I₁ + I_g)S – I_g G = 0

Or (Q + S) I₁ – (Q + S + G) I_g = IS__________________________________________ ... (2)

Multiplying Eq. (1) by (Q+S),

(P +R) (Q + S) I₁ + G (Q + S) I_g = R (Q+S) I   ________________________________ … (3)

Multiplying Eq. (2) by (P + R),
(P + R) (Q + S) I₁ – (P + R)(Q + S + G) I_g = S(P +R) I _________________________ … (4)

Subtracting Eq. (4) from Eq. (3),

[G(Q + S) + (P + R) (Q + S + G)] Ig = [R (Q + S) – S (P + R)] I
. : I_g = (QR – PS) I / [G (P + Q + R + S) + (P + R) (Q + S)]  _____________________… (5)

Condition for balance is i_g = 0 i.e., (QR – PS) I = 0. Since I has a finite value QR – PS = 0.

Thus the condition for the balance of the bridge is
P/Q = R/S ____________________________________________________________ … (6)


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