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1. Identify the above device and give the principle behind it.

2. Obtain the mathematical condition for the galvanometer current to be zero.

3. If the balancing length T obtained fora resistance wire in the arrangement is 40 cm.

Find the new balancing length if the same resistance wire is folded to half its length and connected to the same gap.

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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

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

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