(a) Connect 3 resistors `R_(1),R_(2)` and `R_(3)` in parallel between the
point XY with a battery, a plug key and an ammeter in
series as shown in the figure.
`*`Connect the voltmeter in parallel with these resistors
between the terminals X and Y.
`*` Plug the key and note the current I in the ammeter and
Potential difference VB across the resistors.
`*` Take out the plug from the key. Remove the ammeter and voltmeter from the circuit.
Insert the ammeter in series with the resistor `R_(1).` Let the ammeter reading be `I_(1)`.
Similarly measure the currents through `R_(2)` and `R_(3).` Let these be `I_(2)` and `I_(3)` respectively.
It is observed that the total current I is equal to the sum of the separate currents through
each branch of the combination.
`I= I_(1)+I_(2)+I_(3)`
Let `R_(P)` be the equivalent resistance of the parallel combination of resistors. By applying
Ohm's law to the parallel combination of resistors, we have `I= V/(R_(P))`.
On applying Ohm's law to each resistor, we get
`I_(1) = V/R_(1), ` `I_(2) = V/R_(2),` `I_(3) = V/R_(3)`
Hence `V/R_(P)=V/R_(1)+V/R_(2)+V/R_(3) or 1/R_(P)=1/R_(1)+1/R_(2)+1/R_(3)`
Thus, the reciprocal of the equivalent resistance of a group of resistances joined in
parallel is equal to the sum of the reciprocals of the individual resistances.
(b) `1/R_(eq)=1/R_(1)+1/R_(2)= 1/12+1/12=2/12`
`R_(eq)=6 Omega` `therefore I = V/R_(eq)=6/6=1A`
