the current in a simple series circuit is 5A.when an additional resistance of 2ohms is introduced the current is reduced to 4A.calculate the original circuit.Assume that the applied potential difference is the same in both the cause.
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case 1
V = IR
V = 5R
as I = 5A
okkk
case 2
V = IR'
V = 4 ( R + 2 )
as I reduced to = 4 A
and R' ( new ) = R + 2
V is Dave in both case so
5 R = 4 ( R + 2 )
5 R = 4 R + 8
so R = 8 ohm
V = IR
V = 5R
as I = 5A
okkk
case 2
V = IR'
V = 4 ( R + 2 )
as I reduced to = 4 A
and R' ( new ) = R + 2
V is Dave in both case so
5 R = 4 ( R + 2 )
5 R = 4 R + 8
so R = 8 ohm
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