हिंदी

Two Cells of Emfs 1.5 V and 2.0 V, Having Internal Resistances 0.2 Ω and 0.3 Ω, Respectively, Are Connected in Parallel. Calculate the Emf and Internal Resistance of the Equivalent Cell. - Physics

Advertisements
Advertisements

प्रश्न

Two cells of emfs 1.5 V and 2.0 V,  having internal resistances 0.2 Ω and 0.3 Ω, respectively, are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.

Advertisements

उत्तर

Given:
E1=1.5 V

E2=2 V

r1=0.2 Ω

r2 =0.3 Ω

The effective emf of two cells connected in parallel can be calculated as follows: 

`E_(eff) = (E_1r_2+E_2r_1)/(r_1+r_2)`

`=> E_(eff) = (1.5xx0.3+2.0xx0.2)/0.5 = 1.7 V`

The effective resistance can be calculated as follows:

`R_(eff)= (r_1r_2)/(r_1+r_2)`

`=> R_(eff)= (0.2xx.03)/0.5`0.12 Ω

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2015-2016 (March) Delhi Set 1

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Distinguish between emf and terminal voltage of a cell.


The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?


A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.


In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.


The equivalent resistance between points. a and f of the network shown in Figure 2 is :

a) 24 Ω

b) 110 Ω

c) 140 Ω

d) 200 Ω


The following figure shows an arrangement to measure the emf ε and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed, the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.


A battery of emf 100 V and a resistor of resistance 10 kΩ are joined in series. This system is used as a source to supply current to an external resistance R. If R is not greater than 100 Ω, the current through it is constant up to two significant digits.
Find its value. This is the basic principle of a constant-current source.


Two batteries of emf ε1 and ε22 > ε1) and internal resistances r1 and r2 respectively are connected in parallel as shown in figure.


A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, versus R.


The terminal voltage of the battery, whose emf is 10 V and internal resistance 12, when connected through an external resistance of 40 as shown in the figure is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×