In the circuit diagram above , the ammeter reads a current of 3A when R is 5Ω and 6A when R is 2Ω. Determine the value of x
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Correct Answer: Option C
Explanation:
for a parallel R and X, when R = 5Ω,
and when R = 2Ω
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
for a parallel R and X, when R = 5Ω,
| 5+X |
and when R = 2Ω
| 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