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Point out the right statements about the validity of Kirchhoff's junction rule
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.
The figure below shows currents in a part of electric circuit. The current i is
Applying Kirchhoffs first law,
1 = 2 + 2  I  1.3 = 1.7 A
A circuit has a section ABC if the potential at point A, B and C are V_{1}, V_{2 }and V_{3} respectively, calculate the potential at point O is
Applying juction rule −I_{1 }− I_{2} − I_{3 }= 0
i.e., I_{1 }+ I_{2} + I_{3} = 0
Let, V_{0} bet the potential at point O. By Ohm's law for resistance, R_{1}, R_{2}and R_{3} respectively, we get
So substituting these values of I_{1}, I_{2 }and I_{3} in eq. (i), we get
The potential difference between A and B in figure is
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
In the given circuit the potential at point B is zero, the potential at points A and D will be
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.
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 I_{1} and I_{2} are
From Kirchhoff's first law at jucntion P
I_{1 }+ I_{2} = 6…(i)
From Kirchhoff's second law to the closed circuit PQRP,
−2I_{1} − 2I_{1} + 2I_{2}= 0
⇒ −4I_{1} + 2I_{2 }= 0
⇒ 2I_{1} − I_{2} = 0
Adding Eqs. (i) and (ii), we get
3I_{1} = 6
⇒ I_{1} = 2A
From Eq. (i),
I_{2} = 6 − 2 = 4A
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
Using Kirchoff's law in loop AP_{2}P_{1}DA
∴ 10I_{1} + 2I − 7 = 0
10I_{1} + 2I = 7...(i)
Using Kirchhoff's law in loop P_{2}P_{1}CBP_{2}
−3 + I(I − I_{1}) − 10I_{1 }= 0
I − 11I_{1} = 3, I = 3 + 1_{1}I_{1}....(ii)
From (i) and (ii)
10I_{1} + 2(3 + 11I_{2}) = 710I_{1} + 6 + 22I_{1} = 7
∴ 32I_{1 }= I, I_{1} = 1/32 = 0.031A
In the circuit shown, the value of currents I_{1}, I_{2} and I_{3} are
Applying Kirchhoff's voltage law,
In loop I,
−27 − 6I_{2} − 2I_{1} + 24=0
6I_{2} + 2I_{1 }= −3…(i)
In loop II,
−27 − 6I_{2} + 4I, + 24 = 0
6I_{2} − 4I_{3} = −27…(ii)
At junction P, I_{1} − I_{2} − I_{3} = 0…(iii)
Solving equations (i), (ii) and (iii) we get
I_{1} = 3A, I_{2} = −3/2A, I_{3} = 9/2A.
In the circuit shown, current flowing through 25V cell is
Applying KVL in loop
ABCDA, ABFEA, ABGHA and ABJIA, we get
30 − i_{1} × 11 = −25… (i)
20 + i_{2} × 5 = 25... (ii)
5 − i_{3} × 10 = −25… (iii)
10 + i_{4} × 5 = 25... (iv)
Solving equations (i), (ii), (iii) and (iv) we get
i_{1} = 5A, i_{2} = 1A, i_{3} = 3A and i_{4} = 3A
Hence, current flowing through 25V cell is 12A.
A battery, an open switch and a resistor are connected in series as shown in figure.
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?
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.
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