For a given voltage drop, what would happen to the electric current through the resistance if the value of the resistance was
a.) Doubled
b.) Halved
c.) Five times as large
See suggestion below.
Let's start with the basics.
V = IR
But the form you need is solved for the electric current.
1) Which variable is that?
2) Can you solve the above equation for that variable?
3) One of the variables remains constant for a, b, and c. Which variable is that?
When the value of the resistance changes, the electric current through the resistance will also change according to Ohm's Law, which states that the current (I) flowing through a resistor is directly proportional to the voltage drop (V) across it and inversely proportional to the resistance (R) of the resistor (I = V/R).
a.) If the resistance is doubled:
Using Ohm's Law, if the resistance (R) is doubled, the current (I) will be halved, assuming the voltage drop (V) remains constant. This is because doubling the resistance increases the opposition to the flow of current, resulting in a reduced current through the resistor.
b.) If the resistance is halved:
Similarly, if the resistance (R) is halved, the current (I) will be doubled, assuming the voltage drop (V) remains constant. Halving the resistance decreases the opposition to the flow of current, allowing for a greater current through the resistor.
c.) If the resistance is five times as large:
If the resistance (R) is five times as large, the current (I) will decrease to one-fifth of its original value, assuming the voltage drop (V) remains constant. Increasing the resistance by a factor of five further increases the opposition to the flow of current, resulting in a reduced current through the resistor.
In summary, when the resistance is increased, the current decreases, and when the resistance is decreased, the current increases. This relationship between resistance and current can be understood using Ohm's Law and the inverse relationship between resistance and current in a direct current (DC) circuit.