Two resistors, with resistance `10 Omega` and `15 Omega`, are to be connected to a battery of e.m.f. 12 V so as to obtain :
(i) minimum current (ii) maximum current
(a) Describe the mode of connecting the resistances in each case.
(b) Calculate the strength of the total current in the circuit in each case.

Explanation

(i) Resistances are connected in series to obtain minimum current
`R_(1)=10 Omega" "R_(2)=15 Omega" Voltage"=12 V`
Resultant, `R=R_(1)+R_(2)=10+15=25 Omega`
Potential difference, `V=12 V`
`I=?`

According to Ohm's Law :
`V=IR" ":. I=V/R=12/25=0.48 A`
(ii) The resistances are connected in parallel to obtain maximum current
`R_(1)=10 Omega" "R_(2)=15 Omega" "V=12` Volts
`1/R=1/R_(1)+1/R_(2)" "implies 1/R=1/10+1/15=(3+2)/30=5/30`
`implies 1/R=1/6" "implies R= 6 Omega`
According to Ohm's Law, `V=IR`
`:. I=V/R=12/6=2A`