In the circuit shown below, the ammeter reading is zero. Then, the value of the resistance R is

Options

A. `50 Omega`
B. `100 Omega`
C. `200 Omega`
D. `400 Omega`

Answer

B

Explanation

Given circuit is shown in figure

In loop (1),
`12-500i_(1)-Ri_(1)=0`
`rArr 12=i_(1)(500+R)`
In loop (2).
`12-500i_(1)-2=0rArr10=500i_(1)`
or `i_(1)=(1)/(50)`A
From Eqs. (i) and (ii), we get
`12xx(1)/(i_(1))=(500+R)`
`rArr 12xx50=500+RrArrR=100Omega`