In the circuit shown in figure the reading of ammeter is the same with both switches open as with both closed. Then find the resistance `R`. (ammeter is ideal).

Explanation

Step I: Discuss the circuit when both switches are open:
According to loop rule: `1.5 - 300I - 100I - 50I = 0`
`:. I = 1.5/450 = 15/4500`
`= 1/300A`
Step II: Discuss the circuit
after closing the switch.
In loop ABCDEA,

`-IR + 1.5 - 300I_1 = 0`
or `300I_1 + IR = 1.5`.....(i)
In loop BCGFB,
`-100I +(I_1-I)R = 0`
or `(I_1-I)R = 100I`
or `I_1R = (100+R)I`
`:. I_1 = ((100+R)I)/R`........(ii)
From eqs. (i) and (ii),
`300 ((100+R)I)/R + IR = 1.5`
or ` 300(100+R)/R 1/300 + R/300 = 1.5`
`:. R = 6000Omega`.