In the circuit diagram above, the ammeter reads a current of 3A when R is 5 ohms and 6A when R is 2ohms, the value of the unknown resistance X is
E = I( R + r ) but r = 0 and effective resistance = R + X
E = I(R + X )
E = 3( 5 + X ) .......................I
E = 6(2 + X) ........................II
since E = E
3(5 + X) = 6( 2 + X)
15 + 3X = 12 + 6X
15 - 12 = 6X - 3X
3X = 3
X = 1
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