Advertisements
Advertisements
प्रश्न
State and explain Kirchhoff’s rules.
Advertisements
उत्तर
Kirchhoff’s first rule (current rule or junction rule):
Statement: It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Kirchhoff ’s current rule
Explanation:
All charges that enter a given junction in a circuit must leave that junction since charge cannot build up or disappear at a junction. Current entering the junction is taken as positive and current leaving the junction is taken as negative.
Applying this law to the junction A,
I1 + I2 – I3 – I4 – I5 = o
Or
I1 + I2 = + I3 I4 + I5
Kirchhoff’s second rule (voltage rule or loop rule):
Statement: It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system. (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).
Explanation:
The product of current and resistance is taken as positive when the direction of the current is followed. Suppose if the direction of current is opposite to the direction of the loop, then product of current and voltage across the resistor is negative. It is shown in following Fig. (a) and (b). The emf is considered positive when proceeding from the negative to the positive terminal of the cell. It is shown in following fig. (c) and (d).
(a)
(b)
(c)
(d)
Kirchhoff voltage rule
APPEARS IN
संबंधित प्रश्न
Kirchhoff's voltage law and current law are respectively in accordance with the conservation of .................................. .
- charge and momentum
- charge and energy
- energy and charge
- energy and momentum
State Kirchhoff's rules and explain on what basis they are justified.
Twelve wires, each of equal resistance r, are joined to form a cube, as shown in the figure. Find the equivalent resistance between the diagonally-opposite points a and f.

In the circuit shown in the figure below, E1 and E2 are two cells having emfs 2 V and 3 V respectively, and negligible internal resistance. Applying Kirchhoff’s laws of electrical networks, find the values of currents l1 and I2.

How the emf of two cells are compared using potentiometer?
Figure shows current in a part of an electrical circuit. Then current I is ______.

In a meter bridge the point D is a neutral point (Figure).
- The meter bridge can have no other neutral point for this set of resistances.
- When the jockey contacts a point on meter wire left of D, current flows to B from the wire.
- When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.
- When R is increased, the neutral point shifts to left.
Power P is to be delivered to a device via transmission cables having resistance RC. If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced.
The circuit in figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8Ω. Find the potential difference between the points A and B.
Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2V battery (Figure). Find the effective voltage and effective resistance of the combination.

