(a) With the help of a circuit diagram, obtain the relation for the equivalent resistance of two resistances connected in parallel.

(b) In the circuit diagram shown below, find:

(i) Total resistance.

(ii) Current shown by the ammeter *A *

#### Solution

Let the individual resistance of the two resistors be *R*_{1} and *R*_{2} and their combined resistance be *R*. Let the total current flowing in the circuit be *I* and strength of the battery be *V *volts. Then, from Ohm's law, we have:*V* = *IR* ... (1)

We know that when resistors are connected in parallel, the potential drop across each resistance is the same.

Therefore:*I* = *I*_{1}_{ }+ *I*_{2}*I* = *V*/*R*_{1} + *V*/*R*_{2}*I* = *V*/(1/*R*_{1} + 1/*R*_{2}) ... (2)

From the equations (1) and (2) we have:

1/*R* = 1/*R*_{1} + 1/*R*_{2 }

_{As shown Resistance 2Ω and 3Ω are in senies }

_{We know resistance in series arrangement can be obtained as `R=R_1+R_2` }

_{R=2Ω+3Ω=5Ω }

\[\text{ Since this 5 \Omega is in parallel with another 5 \Omega resistance therefore total resistance can be obtained as }\]

_{`1/R=1/R_1+1/R_2 "Here" R_1=5`Ω `R_2=5`Ω }

_{`1/R=1/5+1/5` }

\[\frac{1}{R} = \frac{2}{5}\]

\[ R = 2 . 5 \Omega\]

`"Total resistance of combination = 2 . 5 \Omega}"`

The current shown by the ammeter *A, *i.e. the current in the circuit can be calculated as:

`I=V/R`

`I=4/2.5=1.6A`