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

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 = I2R).

## 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 = I2R) 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 = I2R can also be used here.

R = P /I2 = 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.

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