the power (P) generated in a electrical circuit is proportional to the square of the current ( L amps) . when the power is 108, the current is 6 amps.
a_ write down the relationship between P,L and the constant of variation k .
b_ calculate the value of k
c_ calculate the value of L when p= 75 watts
P = k L^2
when P=108, L=6
108 = 36K
k = 3
P = 3L^2
75 = 3L^2
L^2 = 25
L = 5
a) The relationship between power (P), current (L), and the constant of variation (k) can be expressed as:
P = k * L^2
b) To calculate the value of k, we can substitute the given values into the equation:
P = k * L^2
108 = k * 6^2
108 = k * 36
To find k, we divide both sides of the equation by 36:
k = 108 / 36
k = 3
So, the value of k is 3.
c) To calculate the value of L when P = 75 watts, we can rearrange the equation:
P = k * L^2
75 = 3 * L^2
To find L, we take the square root of both sides:
√(75 / 3) = L
√25 = L
L = 5
Therefore, when P = 75 watts, the value of L is 5 amps.
a) The relationship between power (P), current (L), and the constant of variation (k) can be expressed as follows:
P = k * L^2
b) To calculate the value of k, we can use the given information that when the power (P) is 108 watts, the current (L) is 6 amps. Plugging these values into the equation, we have:
108 = k * 6^2
108 = k * 36
To solve for k, we divide both sides of the equation by 36:
k = 108 / 36
k = 3
Therefore, the value of k is 3.
c) To calculate the value of L when P = 75 watts, we can rearrange the equation:
P = k * L^2
Substituting the known values, we have:
75 = 3 * L^2
To solve for L, we need to isolate the L term:
75 / 3 = L^2
25 = L^2
Taking the square root of both sides, we find:
L = √(25)
L = 5
Therefore, when P = 75 watts, the value of L is 5 amps.