Advertisements
Advertisements
Question
State Kirchhoff's rules and explain on what basis they are justified.
Solution
Kirchhoff’s first rule (Junction rule): The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.
\[\sum_{} I = 0\]
This law is justified on the basis of law of conservation of charge.
Kirchhoff’s second law (Loop rule): In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and the current flowing through them.
\[\sum_{} I = 0\]
This law is justified on the basis of law of conservation of energy.
APPEARS IN
RELATED QUESTIONS
The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression for
- the current draw from the cell and
- the power consumed in the network.
Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (11/5) Ω?
Determine the equivalent resistance of networks shown in Fig.
An infinite ladder is constructed with 1 Ω and 2 Ω resistors, as shown in the figure. (a) Find the effective resistance between the points A and B. (b) Find the current that passes through the 2 Ω resistor nearest to the battery.
Consider the potentiometer circuit as arranged in the figure. The potentiometer wire is 600 cm long. (a) At what distance from the point A should the jockey touch the wire to get zero deflection in the galvanometer? (b) If the jockey touches the wire at a distance of 560 cm from A, what will be the current in the galvanometer?
Solve the following question.
Using Kirchhoff’s rules, calculate the current through the 40 Ω and 20 Ω resistors in the following circuit.
State Kirchhoff ’s voltage rule.
A copper wire of 10-6 m2 area of cross-section, carries a current of 2 A. If the number of electrons per cubic meter is 8 × 1028, calculate the current density and average drift velocity.
The e.m.f of The battery in a thermocouple is doubled. The rate of heat generated at one of the junction will.
The figure below shows two batteries, E1 and E2, having emfs of 18V and 10V and internal resistances of 1 Ω and 2 Ω, respectively. W1, W2 and W3 are uniform metallic wires AC, FD and BE having resistances of 8 Ω, 6 Ω and 10 Ω respectively. B and E are midpoints of the wires W1 and W2. Using Kirchhoff's laws of electrical circuits, calculate the current flowing in the wire W3: