In the following circuit diagram, the galvanometer reading is zero. If the internal resistance of cells is negligible then the value of x will be
A
10Ω
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B
100Ω
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C
200Ω
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D
500Ω
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Solution
The correct option is B100Ω Applying Kirchhoff's law for outer loop, 12−2=500I⇒I=10500=1/50A As galvanometer reading is zero so the potential across resistance X is IX=2 or, (1/50)X=2⇒X=100Ω