Advertisements
Advertisements
प्रश्न
ε1 and ε2 are two batteries having emf of 34V and 10V respectively and internal resistance of 1Ω and 2Ω respectively. They are connected as shown in the figure below. Using Kirchhoff’s Laws of electrical networks, calculate the currents I1 and I2.

Advertisements
उत्तर
By Kirchoff’s law of electrical networks consider closed loop ABEFA
`:. -4I_1 - 5(I_2 + I_2 ) - 7I_1 - I_1 + 34= 0`
`:. -4I_1 - 5I_1 - 5I_2 - 7I_1 - I_1 = -34`
`:. -17I_1 - 5I_2 = -34`
`:. 17I_1 + 5I_2 = 34` .....(1)
Consider closed loop BCDEB,
`4I_2 - 10 + 2I_2 + 7I_2 + 5(I_1 + I_2) = 0`
`:. 4I_2 + 2I_2 + 7I_2 + 5I_1 + 5I_2 = 10`
`18I_2 + 5I_1 = 10` ....(2)
Mutiply equation (1) by 5 and equation (2) by 17 and Substract

∴ `I_1 = 2A`
APPEARS IN
संबंधित प्रश्न
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.
Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would b the potential difference between points A and B?

The instrument for the accurate measurement of the e.m.f of a cell is ______.
Two cell of 1.25 V and 0.75 V are connected parallel. The effective voltage will be:-
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.
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?
Derive the equation of the balanced state in a Wheatstone bridge using Kirchhoff’s laws.
In a closed loop containing cells and resistors, what equals the total potential drop across the resistors?
While moving opposite to the direction of current through a resistor, how is the potential change treated?
At junction P, currents \[0.2\,\mathrm{A}\], \[0.6\,\mathrm{A}\], and \[0.7\,\mathrm{A}\] point towards P, while \[0.4\,\mathrm{A}\], \[0.5\,\mathrm{A}\], and \[I\] point away. What is the unknown current?
