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ICSE Class 10 Answered

A cell of EMF 2V and internal resistance 1ohm is connected with two wires of resistances 2ohm and 3ohm in parallel. Find the current in each wire.
Asked by www.utkarshdubey1408 | 20 Apr, 2022, 10:52: PM
Expert Answer
Given that,
Internal resistance of cell, r = 1 Ohm
EMF of cell, V = 2 V
Now,
The effective reisistance of wires connected in parallel combination will be given as
1 over R subscript f equals 1 over R subscript 1 plus space 1 over R subscript 2 rightwards double arrow R subscript f space equals fraction numerator R subscript 1 R subscript 2 over denominator R subscript 1 plus R subscript 2 end fraction
therefore space R subscript f equals fraction numerator 3 cross times 2 over denominator 3 plus 2 end fraction equals 6 over 5 equals 1.2 space capital omega space
Hence the amount of current flowing through given combination will be 
I = V(Rf + r) 
equals space fraction numerator 2 over denominator left parenthesis 1.2 plus 1 right parenthesis end fraction equals space 0.9 space space A space
Now, the potential drop across parallel combination of wires will be 
Vf = IRf =0.9 x 1.2 = 1.08 V
Thus,
Current through 3 Ohm resistance will be
I2 = Vf /R2 = 1.08 / 3 = 0.36 A
Similarly, Current through 2 Ohm resistance will be
I1 = Vf /R1 = 1.08 / 2 = 0.54 A
Answered by Jayesh Sah | 23 Apr, 2022, 08:11: PM
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