Dertermine the current through the battery of internal resistance `0.5 Omega` for the circuit shown in fig. `7.14`. How much power is dissipated in `6 omega` resistance ?

Explanation

Resistance of arm `BCDE` is ` 7 + 1+10 = 18 Omega` . Here `18 Omega and 6 Omega` are parallel. Their effective resistance is
`R_(p) = ( 18 xx 16 )/( 18 + 6 ) = ( 18 xx 6 )/( 24) = 4.5 Omega`
Total resistance of the circuit is ` 2+ 4.5 + 8 + 0.5 = 15 Omega` .
Therefore, current through the circuit is
` i = ( 15) / ( 15) = 1 A`
Potential difference across `B and E` is `i xx R_(p) = 1 xx 4.5 = 4.5 V`.
Therefore, power dissipated as heat due to resistance `6 Omega `is
`((4.5)^(2))/(6) = 3.375 W`