Question

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

Answer 1

(a) Since `R_1,R_2` and `R_3` are in parallel, the pd (V) across each one of them is the same, i.e., `12 V`
Thus, `I_1 ("current through" R_1) = (V)/(R_1) = (12 V)/(5 Omega) = 2.4 A`
`I_2 ("current through" R_2) = (V)/(R_2) = (12 V)/(10 Omega) = 1.2 A`
`I_3 ("current through" R_3) = (V)/(R_3) = (12 V)/(30 Omega) = 0.4 A`
(b) total current in the circuit, `I = I_1 + I_2 + I_3 = 2.4 A + 1.2 A + 0.4 A = 4 A`
(c) If `R_p` is the total resistance in the circuit, then
`(1)/(R_p) = (1)/(R_1) + (1)/(R_2) + (1)/(R_3) = (1)/(5) + (1)/(10) + (1)/(30) = (1)/(3)`
or `R_p = 3 Omega`
We can also find the total current by first calculating `R_p` and then using,
`I = (V)/(R_p)`, i.e., `I = (V)/(R_p) = (12 V)/(3 Omega) = 4 A`.