Question

you have a 2Ω, a 3Ω, and a 4Ω resistor in series with a 4.5V battery. Sketch the circuit. What is the total current going through the battery?

Answer 1

Answer:

The sketch of the circuit is shown on the figure 1, where

$$R_1=2\Omega, R_2 = 3\Omega, R_3 = 4\Omega,$$

and

$$E = 4.5V.$$

In series circuit, current remains same in all resistances and battery. The equivalent resistance of the circuit is the next sum

$$R = R_1+R_2+R_3 = 2+3+4 = 9\Omega.$$

The current flows through the battery is

$$I=E/R = 4.5/9 = 0.5A.$$