In the circuit diagram given in Fig. 12.10, suppose the resistors `R_1, R_2 and R_3` have the values `5 Omega , 10 Omega , 30 Omega `, respectively, which have been connected in parallel to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

Explanation

`R_1 = 5 Omega , R_2 = 10 Omega , and R_3 = 30 Omega .`
Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual resistor, therefore, to calculate the current in the resistors, we use Ohm’s law.
The current I1, through `R_1= V// R_1`
`I_1 = 12 V//5 Omega = 2.4 A `
The current `I_2,` through `R_2 = V// R_2`
`I_2 = 12 V//10 Omega = 1.2 A.`
The current `I_3`, through `R_3 = V//R_3 `
`I_3 = 12 V//30 Omega = 0.4 A.`
The total current in the circuit,
` I = I_1 + I__2 + I_3 = (2.4 + 1.2 + 0.4) A `
= 4 A
The total resistance `R_p`, is given by [Eq. (12.18)]
`1/(R_p) = 1/5 +1/10 +1/30 =1/3`
Thus, `R_p = 3 Omega `.