If in Fig. 12.12, `R_1 = 10 Omega , R_2 = 40 Omega , R_3 = 30 Omega , R_4 = 20 Omega , R_5 = 60 Omega `, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Explanation

Suppose we replace the parallel resistors `R_1` and `R_2` by an equivalent resistor of resistance, R ′ . Similarly we replace the parallel resistors `R_3, R_4 and R_5` by an equivalent single resistor of resistance R ″ . Then using Eq. (12.18), we have 1/ R ′ = 1/10 + 1/40 = 5/40, that is R ′ = 8 `Omega `.
Similarly, 1/ R ″ = 1/30 + 1/20 + 1/60 = 6/60,
that is, R ″ = 10 `Omega `
Thus, the total resistance, R = R ′ + R ″ = 18 `Omega `.
To calculate the current, we use Ohm’s law, and get
I = V/R = 12 V/18 `Omega ` = 0.67 A.