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With the help of a labelled diagram, show that the balancing condition of a Wheatstone bridge is ЁЭСЕ1ЁЭСЕ2 =ЁЭСЕ3ЁЭСЕ4 where the terms have their usual meaning.

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With the help of a labelled diagram, show that the balancing condition of a Wheatstone bridge is

`R_1/R_2 = R_3/R_4` where the terms have their usual meaning.

Obtain the balancing condition in the case of Wheatstone’s network.

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Four resistances P, Q, R and S are connected to form a quadrilateral ABCD as shown in the following figure. A battery of emf ε along with a key, is connected between points A and C such that point A is at higher potential with respect to point C. A galvanometer of internal resistance G is connected between points B and D.

When the key is closed, current I flows through the circuit. It divides into I1 and I2 at point A. I1 is the current through P, and I2 is the current through S. The current I1 gets divided at point B. Let Ig be the current flowing through the galvanometer. The currents flowing through Q and R are respectively (I1 – Ig) and (I2 + Ig),

I = I1 + I2       ...(1)

Consider the loop ABDA. Applying Kirchhoff’s voltage law in the clockwise sense shown in the loop, we get

–I1P – IgG + I2S = 0      ...(2)

Applying Kirchhoff's voltage law to loop BCDB in a clockwise sense, we get,

–(I1 – Ig)Q + (I2 + Ig)R + IgG = 0       .....(3) 

From these three equations (Eq. (1), (2), (3), we can find the current flowing through any branch of the circuit.

A special case occurs when the current passing through the galvanometer is zero. In this case, the bridge is said to be balanced. The condition for the balance is Ig = 0. This condition can be obtained by adjusting the values of P, Q, R and S. Substituting Ig = 0 in Eq. (2) and Eq. (3) we get,

–I1P + I2S = 0 ∴ I1P = I2S    ...(4)

–I1Q + I2R = 0 ∴ I1Q = I2R    ...(5)

Dividing Eq. (4) by Eq. (5), we get

`∴ (I_1 P)/(I_1 Q) = (I_1 S)/(I_2 R)`

`therefore P/Q = S/R`

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рдкрд╛рда 9: Current Electricity - Exercises [рдкреГрд╖реНрда реиреирео]

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рдмрд╛рд▓рднрд╛рд░рддреА Physics [English] Standard 12 Maharashtra State Board
рдкрд╛рда 9 Current Electricity
Exercises | Q 3 | рдкреГрд╖реНрда реиреирео

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Choose the correct:

Four resistances 10 Ω, 10 Ω, 10 Ω and 15 Ω form a Wheatstone’s network. What shunt is required across 15 Ω resistor to balance the bridge


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[S = shunt and G = resistance of galvanometer]


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In a Wheatstone bridge, when the potentials at points B and D are the same, then the current through the galvanometer ______

 


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The metre bridge works on the principle of ______.


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In the Wheatstone bridge, (shown in the figure) X = Y and A > B. The direction of the current between a and b will be ____________.


What is the e.m.f of the cell C in the circuit shown in figure, if the deflection in the galvanometer is zero, the resistance of the wire is 3 `Omega`, The length of the wire is 100 cm?


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In the meter bridge experiment, the null point is obtained at a distance of тДУ from the left end. The resistance in the left and right gaps are halved and then interchanged. The new position of the null point is at ______


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