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
Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (11/3) Ω?
Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of 6 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.
In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine the potential at point B.
State Kirchhoff’s current rule.
State Kirchhoff ’s voltage rule.
The e.m.f of The battery in a thermocouple is doubled. The rate of heat generated at one of the junction will.
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?
