Test: Kirchhoff's Laws (NCERT) - NEET MCQ

By Kirchhoff's junction rule. Incoming current = Outgoing current
The net charge is conserved and it is based on conservation of charge.
Also bending or reorienting the wire does not invalidate the conservation of charge principle.

Applying Kirchhoffs first law,
1 = 2 + 2 - I - 1.3 = 1.7 A

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

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.

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

Applying Kirchhoff's voltage law,

In loop I,
−27 − 6I₂ − 2I₁ + 24=0
6I₂ + 2I₁ = −3…(i)
In loop II,
−27 − 6I₂ + 4I, + 24 = 0
6I₂ − 4I₃ = −27…(ii)
At junction P, I₁ − I₂ − I₃ = 0…(iii)
Solving equations (i), (ii) and (iii) we get
I₁ = 3A, I₂ = −3/2A, I₃ = 9/2A.

Applying KVL in loop 
ABCDA, ABFEA, ABGHA and ABJIA, we get
30 − i₁ × 11 = −25… (i)
20 + i₂ × 5 = 25... (ii)
5 − i₃ × 10 = −25… (iii)
10 + i₄ × 5 = 25... (iv)
Solving equations (i), (ii), (iii) and (iv) we get
i₁ = 5A, i₂ = 1A, i₃ = 3A and i₄ = 3A
Hence, current flowing through 25V cell is 12A.

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.