(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 R1 and R2 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 = I1 + I2
I = V/R1 + V/R2
I = V/(1/R1 + 1/R2) ... (2)
From the equations (1) and (2) we have:
1/R = 1/R1 + 1/R2
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`
