A battery of e.m.f. 9 volt. and internal resistance 6 ohm is connected to a parallel combination of two resistors of resistances 20 ohm and 30 ohm. Çalculate the terminal voltage of the battery.

Explanation

E = 9 V
` r = 6 Omega`
`R_1 = 20 Omega`
`R_2=30Omega`
`V= ?`
`:. R_1 and R_2` are in parallel and let their net resistance be R.
`implies 1/R=1/(R_1) +1/R_2`
`= 1/R=1/20 +1/30 = (3+2)/60 = 5/60`
`implies 1/R =1/12`
`implies R = 12 Omega`
Now `r = (E/V -1) R`
`implies 6 = (9/V-1) 12`
`implies 9/V -1 =6/12`
`implies 9/V =1/2 +1 =3/2`
`implies V = (9xx2)/3= 6 ` Volt