NEET Class neet Answered
construction the equivalent cell for following circuit(2V & 5ohm, -3V & 10ohm, 4V & 15ohm)
A) 1V & 30/11ohm
B) 29/30V & 30/11 ohm
![question image](http://images.topperlearning.com/topper/new-ate/top_mob16804136811385678516answer4101724706566647416.jpg)
Asked by dhivi242004 | 02 Apr, 2023, 11:04: AM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/8a1904caf9b1348b9c8907eb19f406166429dd664b9367.19082694f1.png)
To get equivalenet resistance , we short all the voltage sources. If we short all voltage sources , we find thee resistors are in parallel.
Hence equivalenet resistance of circuit is calculated as
![begin mathsize 14px style R subscript e q end subscript space equals space fraction numerator 5 space cross times space 10 space cross times space 15 over denominator 50 plus 150 plus 75 end fraction space equals space 30 over 11 capital omega end style](https://images.topperlearning.com/topper/tinymce/cache/592f4a0e07920d6ab0baa64798a5e3a6.png)
To get the net EMF of the circuit , let us apply Kirchoff's rules.
Let us assume the current distribution as shown in figure .
If we apply kirchoff's voltage law to the loop ABCFA, we get
-2 -5 I3 + 10 I2 - 3 = 0
After simplification of above expression , 2 I2 - I3 = 1 . . . . . . . . . . . . . . . . . . . . . (1)
If we apply kirchoff's voltage law to the loop FCDEF , we get
3 - 10 I2 - 15 I1 + 4v = 0
2 I2 + 3 I1 = 7/5 .. . . . . . . . . . . . . . . . . . . (2)
If we apply kirchoff's current rule at node C , we get , I1 = I2 + I3 . . . . . . . . . . . . . . . . . (3)
After substituting for I1 using eqn.(3) , we rewrite eqn.(2) after simplification as
5 I2 + 3 I3 = 7/5 . . . . . . . . . . . . . . . . . . . . . ( 4 )
By solving equations (1) and (5), we get I2 = ( 2/5 ) A and I3 = ( -1/5 ) A
BY substituting I2 and I3 from eqn.(3) , we get I1 = ( 1/5 ) A
Potential difference across E and D or Potential difference across F and C or Potential difference across A and B is equivalent EMF E of circuit
we get E = - 1 V
Equivalent cell of given circuit is 1 V and (30/11) Ω
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