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Test: Kirchhof's Laws (NCERT)


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Test: Kirchhof's Laws (NCERT) - Question 1

Point out the right statements about the validity of Kirchhoff's junction rule

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 1

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.

Test: Kirchhof's Laws (NCERT) - Question 2

The figure below shows currents in a part of electric circuit. The current i is

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 2

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

Test: Kirchhof's Laws (NCERT) - Question 3

A circuit has a section ABC if the potential at point A, B and C are V1, Vand V3 respectively, calculate the potential at point O is

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 3

Applying juction rule −I− I2 − I= 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, Iand I3 in eq. (i), we get


Test: Kirchhof's Laws (NCERT) - Question 4

The potential difference between A and B in figure is

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 4

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 VC − VA = (2Ω)(1A) = 2V...(i)
Potential difference across C and B is VC − VB = (3Ω)(1A) = 3V...(ii)
Substracting (i) from (ii), we get
VA − VB = 3V − 2V = 1V

Test: Kirchhof's Laws (NCERT) - Question 5

In the given circuit the potential at point B is zero, the potential at points A and D will be
Physics Question Image

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 5

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

Test: Kirchhof's Laws (NCERT) - Question 6

A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2 Ω each and leaves by the corner R. Then the current I1 and I2 are

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 6

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

Test: Kirchhof's Laws (NCERT) - Question 7

A 7V battery with internal resistance 3Ω and 3V battery with internal resistance 10mega are connected to a 10Ω resistors as shown in figure, the current in 10Ω resistor is

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 7

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) − 10I= 0

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

Test: Kirchhof's Laws (NCERT) - Question 8

In the circuit shown, the value of currents I1, I2 and I3 are

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 8

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.

Test: Kirchhof's Laws (NCERT) - Question 9

In the circuit shown, current flowing through 25V cell is

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 9

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.

Test: Kirchhof's Laws (NCERT) - Question 10

A battery, an open switch and a resistor are connected in series as shown in figure.
image
Consider the following three statements concerning the circuit. A voltmeter will read zero if it is connected across points
(i) P and T
(ii) P and Q
(iii) Q and T
Which one of the above is/are true?

Detailed Solution for Test: Kirchhof's Laws (NCERT) - Question 10

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

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.

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