Test: Kirchhoff’s Rules & Wheatstone Bridge (December 6) - NEET MCQ

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

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

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

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., V_B − V_D = 0 volt

Applying juction rule −I₁ − I₂ − I₃ = 0
i.e., I₁ + I₂ + I₃ = 0
Let, V₀ bet the potential at point O. By Ohm's law for resistance, R₁, R₂and R₃ respectively, we get

So substituting these values of I₁, I₂ and I₃ in eq. (i), we get

Resistance of the upper arm CAD = 2Ω + 3Ω = 5Ω
Resistance of the lower arm CBD = 3Ω + 2Ω = 5Ω
As the resistance of both arm are equal, therefore same amount of current flows in both the arms. Current through each arm. CAD or CBD = 1A
Potential difference across C and A is V_C − V_A = (2Ω)(1A) = 2V...(i)
Potential difference across C and B is V_C − V_B = (3Ω)(1A) = 3V...(ii)
Substracting (i) from (ii), we get
V_A − V_B = 3V − 2V = 1V

From Kirchhoff's first law at jucntion P 
I₁ + I₂ = 6…(i)
From Kirchhoff's second law to the closed circuit PQRP,
−2I₁ − 2I₁ + 2I₂= 0
⇒ −4I₁ + 2I₂ = 0
⇒ 2I₁ − I₂ = 0
Adding Eqs. (i) and (ii), we get
3I₁ = 6
⇒ I₁ = 2A
From Eq. (i),
I₂ = 6 − 2 = 4A

Using Kirchoff's law in loop AP₂P₁DA
∴ 10I₁ + 2I − 7 = 0
10I₁ + 2I = 7...(i)
Using Kirchhoff's law in loop P₂P₁CBP₂
−3 + I(I − I₁) − 10I₁ = 0

I − 11I₁ = 3, I = 3 + 1₁I₁....(ii)
From (i) and (ii)
10I₁ + 2(3 + 11I₂) = 710I₁ + 6 + 22I₁ = 7
∴ 32I₁ = I, I₁ = 1/32 = 0.031A

V_A − V_B = 2 × 2 = 4V
∴ V_A − 0 = 4V
⇒ V_A = 4V
According to question V_B = 0
Point D is connected to positive terminal of battery of emf 3V.

When the switch is not closed, a voltmeter connected across P and T will not show any potential difference.

 

Between Q and T also there is no potential difference because circuit is not complete.
Therefore in both the cases, the voltmeter will read zero. Between P and Q, the emf of the battery will be given.