In the circuit shown, the variable resistance X is to be adjusted such that the ideal ammeter reads the same in both the positions of the key,when connected independently to 1 and then to 2. The reading of the ammeter is 2 A. If E = 10 V then X is
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a
5 Ω
b
20 Ω
c
50 Ω
d
cannot be determined
answer is A.
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Detailed Solution
In position (1) : ε−ir=0 ⇒ ε=2rNow, in position (2): ε−ir+E−iX=0 ⇒ 2r−2r+10−2X=0 ⇒ X=5Ω