In the circuit diagram above , the ammeter reads a current of 3A when R…

Explanation

for a parallel R and X, when R = 5Ω,

effective resistance R =	5X
	5+X

and when R = 2Ω

effetive resistance R =	2X
	2+X

Thus since E = I(R+r), and if r = 0 then E = IR
∴(i) E = 3(5X/(5+x)); (ii) E = 6(2X(2+x))
∴3(5X/(5+X)) = 6(2X/(2+X))
=> (5X/(5+X)) = 2(2x/(2+X))
∴ (5X/(5+X)) = (4X/(2+X))
5x(2+x) = 4x(5+x)
10x + 5x² = 20x + 4x²
5x² - 4x² = 20x - 10x
=>x² = 10x
∴x² - 10x = 0
x(x - 10) = 0
∴ x = 0 or 10
∴ x = 10Ω

There is an explanation video available .