Question

The emf of the battery shown in the figure is 12 V.true or false,explain?​

Answer 1

Answer: True

Explanation:

1)

Firstly,  

As provided in Fig. 1 , set the ground .

Then,  

Using Ohm 's Law ,

Current in 1 ohm resistor is 1 A.

Potential at point A  is

is V = 1*2=2V .

Two 1 ohm resistors are in series , equivalent of the two is 1+1 = 2ohm .

2) Potential across the other 1 ohm resistor is also 2 V,  

So,  similarly Current there will be 1 A.

This means AB resistor 2ohm ,carries 1+1=2A current ( By Kirchoffs current Law)  .

3) Our next manipulated Fig :2 ,

Potential at point B,  is

V-2= 2*2

=> V= 6 volts

4) 2ohm and 1ohm in fig 2 are in series,  

So net will be (2+1)=3ohm .

Also,  3ohm and 6ohm are in parallel,  

That is,  net will be

1/R= 1/3+1/6

=> R = 2 ohm

5) In fig. 3 ,

Current across 2ohm resistor will be

I = (6-0)/2=3A.

Finally,  

Using ohms law,  

E/(2+2)= 3A

=> E = 12V

Hence,  EMF of cell be 12 V.