A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. (a) What is the current through the 4.0-Ω resistor?

The 4ohm resistor is not parallel to any devices, so the current through it is the same as the total current through the circuit.

Apply I = V/R

I = total current, V = battery voltage, R = total resistance

Calculate the total resistance R of the circuit first.

R will be the sum of the 4ohm resistor and the resistance of the 5&9ohm parallel resistors. To find the resistance of the pair, you apply the "sum of inverses" rule:

R = 1 / (1/5 + 1/9) = 3.21ohm

Now add the 4ohm resistor:

R = 3.21 + 4 = 7.21ohm

Now let's calculate I = V/R:

Given values:

V = 6.0V

R = 7.21ohm

Plug in and solve for I:

I = 6.0/7.21

I = 0.83A