# p=iv practice problems | p=iv problems using electric power formula

Last updated on May 6th, 2023 at 05:32 am

In this post on numerical solving, we will see how to solve a set of **p=iv practice problems** or **p=iv problems** using the electric power formulas (P = IV and P = I^{2}R).

**Formulas Used**

**Electric Power Formulas** used to solve the numerical problems listed on this page are as follows:

P = IV

P = I

^{2}R

## p=iv practice problems | Electric power formula p=iv based numerical problems

The **Electric Power formula**-based numerical set is given below with solutions.

1 ) Find the current and resistance of a 60-W, 120-V light bulb in operation.

*approach to solve this problem*

Power = current x voltage, (P = IV), so given P and V the current I may be found.

Knowing the current and the voltage, the resistance could be found (V = IR, Ohm’s law). Also knowing the current, the relationship power = (current)^{2} x resistance (P = I^{2}R) could be used.

**Solution**

Step 1

Given: P = 60 W (power)

V = 120 V (voltage)

Step 2 Wanted: I (current) =?

R (resistance)=?

Step 3 The current is obtained using Eq. P = IV.

Rearranging yields, I = P/V = 60/120 amp = 0.5 amp

Ohm’s law can be rearranged to get resistance.

R = V/I = 120/0.5 ohm = 240 ohm

An alternate way to find R: Note that the equation P = I^{2}R can also be used here.

R = P /I^{2} = 60/ (0.5)^{2} = 240 ohm

2 ) A coffeemaker draws 10 A of current operating at 120 V. How much electrical energy does the coffeemaker use each second?

Given: I = 10 A

V = 120 V

t = 1 second

Electrical Energy used each second = VIt = 120x10x1 = 1200 Joule.