Question

two cells of EMF 2 volt and 1 volt have internal resistance 1 ohm and 2 ohm respectively .their positive terminal are joined by wire of resistence 10 ohm and negitive terminal by 4 ohm .another wire resistence 10 ohm is connected between middle point of these wire . Find the current through cell and potential difference across 10 ohm wire.​

Answer 1

For parallel combination,

For parallel combination,Net emf, ε=r1+r2ε1r2+ε2r2

For parallel combination,Net emf, ε=r1+r2ε1r2+ε2r2Net internal resistance ,rint=r1+r2r1r2

For parallel combination,Net emf, ε=r1+r2ε1r2+ε2r2Net internal resistance ,rint=r1+r2r1r2For series combination,

For parallel combination,Net emf, ε=r1+r2ε1r2+ε2r2Net internal resistance ,rint=r1+r2r1r2For series combination,Net emf, ε=ε1+ε2

For parallel combination,Net emf, ε=r1+r2ε1r2+ε2r2Net internal resistance ,rint=r1+r2r1r2For series combination,Net emf, ε=ε1+ε2Net internal resistance =rint=r1+r2

For parallel combination,Net emf, ε=r1+r2ε1r2+ε2r2Net internal resistance ,rint=r1+r2r1r2For series combination,Net emf, ε=ε1+ε2Net internal resistance =rint=r1+r2Given ε1=1V,ε2=2V and r1=2Ω,r2=1Ω,Rext=R

For parallel combination,Net emf, ε=r1+r2ε1r2+ε2r2Net internal resistance ,rint=r1+r2r1r2For series combination,Net emf, ε=ε1+ε2Net internal resistance =rint=r1+r2Given ε1=1V,ε2=2V and r1=2Ω,r2=1Ω,Rext=R∴ Current I1=r1+r2+Rε1+ε2=2+1+r1+2

hope it helps uu