if the resistance is halved and the voltage is not changed, what will happen to the current in resistive circuit?
V=IR
When V=constant,
V=(2/2)*IR
=>V=(2*I)*(R/2)
So, double current flow when R is halved.
Now,
P=VI
=>(2*P)=V*(2*I)
So, power double if V=constant & R is halved.
I=E/R
so, the new I will be
E/(R/2) = 2E/R
If the resistance in a resistive circuit is halved and the voltage is not changed, the current in the circuit will double.
To determine what will happen to the current in a resistive circuit if the resistance is halved and the voltage remains unchanged, you can use Ohm's Law. Ohm's Law states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R), or I = V/R.
In this scenario, if the resistance (R) is halved while the voltage (V) remains constant, we can plug in the values to see the effect on the current (I). Let's say the original resistance is R1 and the new resistance is R2, and the voltage is V.
Using Ohm's Law, the original current (I1) can be calculated as:
I1 = V / R1
And the new current (I2) can be calculated as:
I2 = V / R2
Since the voltage (V) is unchanged, we can simplify the equation for I2:
I2 = V / (R1/2) [since R2 = R1/2]
To simplify further, we can multiply the numerator and denominator by 2:
I2 = 2V / R1
Comparing the new current (I2) to the original current (I1), we can see that if the resistance is halved and the voltage remains constant, the current will double. This means that the current in the resistive circuit will increase.