1. Meter bridge.
2. We get galvanometer current as zero, when \(\frac{P}{Q}\) = \(\frac{R}{S}\). For derivation of \(\frac{P}{Q}\) = \(\frac{R}{S}\) Wheat stone’s
Bridge:
Four resistances P, Q, R, and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I1 , I2 , I3 and I4 be the four currents passing through P, R, Q, and S respectively.
Working:
The voltage across R
When key is closed, current flows in different branches as shown in figure.
Under this situation
The voltage across P, VAB = I1 P
The voltage across Q, VBC = I3 Q ……(1)
The voltage across R, VAD = I2 R
The voltage across S, VDC = I1 S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I1 = I3 and I2 = I4 ……(2)
Using Kirchoff’s second law in loop ABDA and BCDB, we get
VAB = VAD ……….(3)
and VBC = VDC…….. (4)
Substituting the values from eq(1) into (3) and (4), we get
I1P = I2R……….(5)
and I3Q = I4S……..(6)
Dividing Eq(5) by Eq(6)
\(\frac{I_3P}{I_3Q}=\frac{I_2R}{I_4S}\)
\(\frac{P}{Q}=\frac{R}{S}\) [since I1 = I3 and I2 = I4]
This is called Wheat stone condition.
3. When we apply this condition in meter bridge, we get
\(\frac{x}{R}=\frac{lr}{(100-lr)}\) x = \(\frac{Rl}{(100-l)}= \frac{R40}{60}=\frac{2R}{3}\)
If wire is folded, New resistance x1 = \(\frac{x}{2}\)
Substituting this in \(\frac{P}{Q}\) = \(\frac{R}{S}\) we get