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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
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Asked by dhivi242004 | 02 Apr, 2023, 11:04: AM
answered-by-expert Expert Answer
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
 
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 I+ 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 I= ( 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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Answered by Thiyagarajan K | 03 Apr, 2023, 02:29: AM
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