Two resistors `4 Omega and 6 Omega` connected in parallel. The combination is connected across a `6 V` battery of negligible resistance. Calculate
(a) the current through the battery
(b) current through each resistor.

Answer

(a) `2.5 A`
(b) `1.5 A, 1.0 A`

Explanation

If `R` is the resultant resistance of `4 Omega and 6 Omega` in parallel, then `R = (4 xx 6)/(4 + 6) Omega = 2.4 Omega`
Current through the battery, `I = (V)/( R)=(6 V)/(2.4 Omega) = 2.5 A`
Since the resistors are in parallel, the pd across each is the same, i.e., `V`
Current through `4 Omega` resistors `= (V)/(4 Omega)=(6 V)/(4 Omega) = 1.5 A`
and current through `6 Omega` resistance `= (V)/(6 Omega) = (6 V)/(6 Omega) = 1.0 A`.