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 `E = 10 V,` then `X` is:

Options

A. `5 Omega`
B. `20 Omega`
C. `50 Omega`
D. cannot be determined

Answer

A

Explanation

(a) In position `(1) epsilon = ir + E = ix = 0 implies epsilon = 2r`.
now, in position `(2) epsilon - ir + E - ix = 0`
`implies 2r = 2r + 10 - 2x = 0 implies x = 5 Omega`