In the circuit shownin figure `V_1` and `V_2` are two voltmeter of resistances `3000 Omega` and ` 2000 Omega` respectively. In additions `R_1=2000Omega`, `R_2=3000omega` and `E=200 V` then
a. Fid the reading of voltmeters `V_1` and `V_2` when
i. switch `S` is open
ii. Switch `S` is closed
b. Current through `S`, when it is closed (Disregard the resistance of battery)

Answer

A, B

Explanation

a. i. When switch `S` is open `V_1 and V_2` are in series, connected to `200 V` battery. Potential will drop in direct ratio of their resitors.
`:. V_1:V_2=R_(V_1):R_(V_2)=3000:2000`
`=3:2`
`:. V_1=3/5xx200=120V`
`V_2 = 2/5xx200=80V`
ii. When `S` is closed then `V_1` and `R_1` are in parallel.Similarly, `V_2` and `R_2` are also in parallel. Now, they are in series and they come out to be equal. So `200 V` will equalilty distribute between them.
`:. V_1=V_2=200/2=100V` each
b. `i_2=100/2000=1/20A`
`i_4=100/3000=1/30A`

If we apply junction law at `P` then current through switch
`=i_2-i_=1/60 A ` in upward direction.