resistence

Asked by RITURAJ KUMAR | 9th Oct, 2010, 09:14: AM

Expert Answer:

Dear Student, 
 
(1) refer to the following circuit diagram which shows the cell with emf E and internal resistance r:
 
In series, Net R = 2 x 2 = 4 ohm
voltage drop across net resistance nad hence across the cell, V = iR = 0.4 x 4 = 1.6
Also, V = Emf - ir         (where EMf: cell emf, r: internal cell resistance)
       1.6 = Emf - (0.4r)...........................................[1]
 
 
 
In parallel, Net R = (2 x 2)/(2+2) = 1 ohm
voltage drop across net resistance and hence across the cell, V = iR = 1.2 x 1 = 1.2
Also, V = Emf - ir         (where EMf: cell emf, r: internal cell resistance)
       1.2 = Emf - (1.2r)............................................[2]
 
From [1] and [2], Internal resistance r = 0.5 ohm
 
 
(2)
 
Net parallel resistance = (3*6)/(3+6) = 2 ohm
 
(i) current flowing through the circuit: i
Then,    Emf = (Rparallel + internal R)i = (3 + 2)i
            15 = 5i
            i = 3 A
 
(ii) p.d between terminals of battery, V = emf - ir = 15 - (3*3) = 6 V
 
(iii) Current in 3 ohm resistor:
 
     p.d. across parallel resistors = p.d. across terminals of battery = 6 V
     6 V = i R = 3 * i
     i = 6/3 = 2 A
 
(iv) current in 6 ohm resistor
 
     p.d. across parallel resistors = p.d. across terminals of battery = 6 V
     6 V = i R = 6 * i
     i = 6/6 = 1 A
 
Hope that satisfies your query.
 
Regards
Team
TopperLearning

Answered by  | 9th Oct, 2010, 06:41: PM

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