Question

A 2.0-ohm resistor is connected in a series with a 20.0 -V battery and a three-branch parallel network with branches whose resistance are 8.0 ohms each. Ignoring the battery’s internal resistance, what is the current in the batter? Show your work.

Answer 1

First of all, let's calculate the equivalent resistance of the three resistors in parallel:

`(1)/(R_p)= (1)/(8 \Omega)+ (1)/(8 \Omega)+ (1)/(8 \Omega)= (3)/(8 \Omega)`

`R_p= (8)/(3) \Omega=2.67 \Omega`

The 2.0-ohm resistance is in series with these resistors, so the equivalent resistance of the circuit is

`R_(eq)=2.0 \Omega + R_p = 2.0 \Omega + 2.67 \Omega = 4.67 \Omega`

And the current in the circuit can now be calculated by using Ohm's law:

`I= (V)/(R_(eq))= (20.0 V)/(4.67 \Omega)=4.28 A`