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# Test: Wheatstone Bridge (NCERT)

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Test: Wheatstone Bridge (NCERT) - Question 1

### In a wheatstone bridge if the battery and galvanometer are interchanged then the deflection in galvanometer will

Detailed Solution for Test: Wheatstone Bridge (NCERT) - Question 1

The deflection in galvanometer will not be changed due to interchange of battery and the galvanometer.

Test: Wheatstone Bridge (NCERT) - Question 2

### In a Wheatstone’s network, P = 2Ω, Q = 2Ω, R = 2Ω and S = 3Ω. The resistance with which S is to be shunted in order that the bridge may be balanced is

Detailed Solution for Test: Wheatstone Bridge (NCERT) - Question 2

Let x be the resistance shunted with S for the bridge to be balanced.
For a balance Wheatstone's bridge

or S' = 2Ω
From Figure

x = 6Ω

Test: Wheatstone Bridge (NCERT) - Question 3

### Four resistances of 3Ω, 3Ω, 3Ω and 4Ω respectively are used to form a Wheatstone bridge. The 4Ω resistance is short circuited with a resistance R in order to get bridge balanced. The value of R will be

Detailed Solution for Test: Wheatstone Bridge (NCERT) - Question 3

The bridge will be balanced when the shunted resistance is of the value of 3Ω

Test: Wheatstone Bridge (NCERT) - Question 4

Four resistors are connected as shown in the figure.

A 6V battery of negligible resistance, is connected across terminals A and C. The potential difference across terminals B and D will be

Detailed Solution for Test: Wheatstone Bridge (NCERT) - Question 4

The given figure is a circuit of balanced Wheatstone bridge as shown in the figure.

Points B and D would be at the same potential,
i.e., VB − VD = 0 volt

Test: Wheatstone Bridge (NCERT) - Question 5

Resistances P, Q, S and R are arranged in a cyclic order to form a balanced Wheatstone's network. The ratio of power consumed in the branches (P + Q) and (R + S) is

Detailed Solution for Test: Wheatstone Bridge (NCERT) - Question 5

For balanced Wheatstone's bridge
P/Q = R/S.....(i)
Power dissipation in resistance R with voltage V is V2/R.
∴
From equation (i)

⇒
Using (i), we get

∴

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