CBSE Class 10 Answered
resistence
Asked by RITURAJ KUMAR | 09 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 | 09 Oct, 2010, 06:41: PM
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