Find the current through the resistance R in Fig if `R =10 Omega` and `R=20Omega`.

Explanation

The p-n junction is reverse biased by battery of e.m.f. 6V. Hence no current through p-n junction. In remaining circuit, both the battries are in series along with resistances R and `10Omega`.
Current through `R(=10Omega)`
`=(6+4)/(10+10)=10/20=1/2=0.5A`
Current through `R(=20Omega)`
`(6+4)/(10+20)=10/30=0.33A`