In an electrical circuit, the current passing through a conductor varies inversely with the resistance. Suppose that when the current is 8A (amperes), the resistance is 15 ohms. What is the current when the resistance is 4 ohms?
That is correct. (15/4)*8 = 30 A
To find the current when the resistance is 4 ohms, we can use the concept of inverse variation. Inverse variation occurs when two variables are related such that their product is constant. In this case, the current and resistance are inversely proportional, meaning their product will always be the same.
First, let's find the constant of variation. Given that when the current is 8A, the resistance is 15 ohms, we can set up the equation:
Current × Resistance = Constant
8A × 15 ohms = Constant
Solving for the constant, we get:
Constant = 120
Now, we can use the constant of variation to find the current when the resistance is 4 ohms. Let's call this current I:
I × 4 ohms = 120
Simplifying the equation, we get:
4I = 120
Dividing both sides of the equation by 4, we find:
I = 120/4
I = 30
Therefore, when the resistance is 4 ohms, the current passing through the conductor is 30A.
To find the current when the resistance is 4 ohms, we can use the concept of inverse variation. Inverse variation means that two quantities are related in such a way that when one increases, the other decreases, and vice versa, while their product remains constant.
In this case, we are given that the current (I) varies inversely with the resistance (R). Mathematically, this can be represented as:
I ∝ 1/R
To find the constant of variation (k), we can use the given information that when the current (I) is 8A, the resistance (R) is 15 ohms. Substituting these values into the equation, we get:
8 ∝ 1/15
8 = k/15
To solve for k, we can multiply both sides of the equation by 15:
8 * 15 = k
k = 120
Now that we have the constant of variation, we can use it to find the current (I) when the resistance (R) is 4 ohms. Substituting the values into the equation:
I = k/R
I = 120/4
I = 30
Therefore, when the resistance is 4 ohms, the current passing through the conductor will be 30 amperes.
E=IR
E=15*8
E=120V
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I=E/R
I=120/4
I=30 amperes