A two-slice bread toaster consumes 849.0 W of power when plugged into a 120.0-V source.
(a) What is the current in the toaster?
A
(b) What is the resistance of the coils in the toaster?
Ω
Given P=849.0, V=120.0, I=? and R=?
Using p=IV and I=P/V
I=849.0/120.0
I=7.075A
Putting I= 7.075 into v=it
120.0=7.075xrR
R=120.0/7.075
R=16.96 ohms
V=IR....not it
a. E*I = 849
120*I = 849
I = 7.075A.
b. R = E/I = 120/7.075 = 16.96 ohms.
To find the current in the toaster, you can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). The formula can be written as:
I = V / R
(a) What is the current in the toaster?
We are given that the toaster is plugged into a 120.0-V source and consumes 849.0 W of power. Power (P) is the product of voltage and current, so we can use this relationship to find the current.
P = V * I
Rearranging the equation, we have:
I = P / V
Substituting the given values, we get:
I = 849.0 W / 120.0 V
Calculating this, we find:
I ≈ 7.075 A
Therefore, the current in the toaster is approximately 7.075 Amperes.
(b) What is the resistance of the coils in the toaster?
To find the resistance (R) of the coils in the toaster, we can rearrange Ohm's Law equation:
R = V / I
Substituting the given values, we have:
R = 120.0 V / 7.075 A
Calculating this, we find:
R ≈ 16.96 Ω
Therefore, the resistance of the coils in the toaster is approximately 16.96 Ohms.