A student was trying to construct the circuit shown in the figure below maked (a), but ended up constructing the circuit marked (b). Realizing her mistake, she corrected the circuit , but to her surprise, the output voltage (across R) did not change.

The value of resistance R is :-

Options

A. 100 `Omega`
B. `150Omega`
C. `200Omega`
D. `300Omega`

Answer

A

Explanation

For circuit (a),
`i_R=(10/((300R)/(300+R)+300)xx300/(300+R))`
Current through cell
[Note : 300 `Omega` & R are in parallel which is in series with `100&200Omega`]
`thereforeV_(R_a)=(10xx300R)/(300R+300^2+300R)`
[`V_(R_a)` is potential difference across resistance R]
For circuit (b),
`i_R=(10/((200+R)(300))/(200+R+300)+100)xx300/(300+200+R)`
Current through cell
[Note : R & 200 `Omega` are in series which is in parallel with 300 `Omega` & again the combination is in series with 100 `Omega`]
`thereforeV_(R_b)=(100xx300R)/(300xx200xx300R+100xx500+100R)`
[ `V_(R_b)` is potential difference across resistance R]
According to given situtaion
`V_(R_a)=V_(R_b)`
`therefore300 R + 9 × 10^4 + 300 R = 6 xx 10^4 + 400 R +5xx10^4`
`rArr200R=2xx10^4rArrR=100Omega`