In the circuit shown below the cells E1 and E2
have emf’s 4 V and 8 V and internal resistance 0.5 ohm and 1 ohm respectively.
Then what will be the potential difference across cell E1 and E2?
Show <=Solution
\[In \ the \ given \ circuit \ external \ resistance \ R=\frac{3\times 6}{3+6}+4.5 =6.5\Omega.\]
\[\therefore \ main \ current \ i=\frac{E_{1}-E_{2}}{R+r_{eq}}=\frac{8-4}{6.5-0.5+0.5}=0.5 A\]
\[E_{1} \ is \ charging \ thus \ E_{1}=V_{1}-ir_{1}\Rightarrow V_{1}=4+0.5\times0.5=4.25\ V\]
\[E_{2} \ is \ discharging \ thus \ E_{2}=V_{2}+ir_{2}\Rightarrow V_{2}=8-0.5\times0.5=7.75\ V\]
