Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
State the two Kirchhoff’s rules used in electric networks. How are there rules justified?
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in the figure. Each resistor has 1 Ω resistance.
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.
Determine the equivalent resistance of networks shown in Fig.
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?
Kirchhoff’s second law is a consequence of law of conservation of ______.
Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The ratio `i_3/i_1` of currents in terms of resistances used in the circuit is:
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.
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:
