Question

A series combination of two resistors, 7.25 ω and 4.03 ω, is connected to a 9.00 v battery.

a. calculate the equivalent resistance of the circuit and the current.

b. what is the potential difference across each resistor?

Answer 1

a.
`11.28\Omega`

The equivalent resistance of a series combination of two resistors is equal to the sum of the individual resistances:

`R_(eq)=R_1 + R_2`

In this circuit, we have

`R_1 = 7.25 \Omega\\R_2 = 4.03 \Omega`

Therefore, the equivalent resistance is

`R_(eq)=7.25 \Omega + 4.03 \Omega=11.28 \Omega`

b. 5.8 V, 3.2 V

First of all, we need to determine the current flowing through each resistor, which is given by Ohm's law:

`I=(V)/(R_(eq))`

where V = 9.00 V and
`R_(eq)=11.28 \Omega`. Substituting,

`I=(9.00 V)/(11.28 \Omega)=0.8 A`

Now we can calculate the potential difference across each resistor by using Ohm's law again:

`V_1 = I R_1 = (0.8 A)(7.25 \Omega)=5.8 V`

`V_2 = I R_2 = (0.8 A)(4.03 \Omega)=3.2 V`