Advertisements
Advertisements
Question
Using Kirchhoff’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Ω resistance. Also find the potential difference between A and D.
Advertisements
Solution
Apply Kirchhoff’s In loop ABCFA:-
I + I + 4I1 = 9 − 6
2I + 4I1 = 3 … (1)
As there is no current flowing through the 4Ω resistance,
I1 = 0
Or, 2I = 3
Or, I = 1.5A
Thus the current through resistances R is 1.5A.
As there is no current through branch CF, thus equivalent circuit will be,
By applying Kirchhoff’s loop law we get,
1.5 + 1.5 + R (1.5) = 9 − 3
R = 2Ω
Potential difference between A and D = (9 − 3) = 6V
APPEARS IN
RELATED QUESTIONS
Kirchhoff's junction law is equivalent to .............................
(a) conservation of energy.
(b) conservation of charge
(c) conservation of electric potential
(d) conservation of electric flux
Use Kirchhoff's rules to obtain conditions for the balance condition in a Wheatstone bridge.
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.
State Kirchhoff's rules for an electric network. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.
Consider the circuit shown in the figure. Find (a) the current in the circuit (b) the potential drop across the 5 Ω resistor (c) the potential drop across the 10 Ω resistor (d) Answer the parts (a), (b) and (c) with reference to the figure.
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.
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?
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.
State the principle of potentiometer.
Obtain the condition for bridge balance in Wheatstone’s bridge.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
Two cell of 1.25 V and 0.75 V are connected parallel. The effective voltage will be:-
Kirchhoff s second law is based on the law of conservation of ______
What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate `R_(unknown)` by any other method?
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.
The value of current in the 6Ω resistance is ______.
