1. If the voltage impressed across a circuit is constant but the resistance increases by a factor of 15. By what factor does the current change?
2. If the resistance of a circuit remains constant while the voltage across the circuit decreases by a factor of 12, by what factor does the current change?
3.A diode converts AC to pulsed DC. What electric device smooths that pulsed DC to a smoother DC?
1.Increasing the resistance by a factor of 15 means multiplying the resistance
by 15. Since the current is inversely
proportional to the resistance,when the resistance is multiplied by 15,
the current is divided by 15. So the
current is decreased by a factor of 15.
2. Decreasing the voltage by a factor of 12 means dividing it by 12.Since the current is directly proportional to the voltage,it is also divided by 12. In other words, the current is
decreased by a factor of 12 also.
Mathematically,Current(I) = V/R.
3. A capacitor (also called a condenser)
is decreased
1. To determine the change in current, we can use Ohm's Law. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance.
Since the voltage across the circuit is constant, let's represent it as V1. Let's also represent the initial current and resistance as I1 and R1, respectively.
According to Ohm's Law, we have V1 = I1 * R1.
If the resistance increases by a factor of 15, we can denote the new resistance as R2 = 15 * R1.
Substituting this into Ohm's Law, we get V1 = I1 * (15 * R1).
We can simplify this equation by canceling out R1 and rearranging to solve for the current:
I1 = V1 / (15 * R1)
So, the current remains the same, I1, when the resistance increases by a factor of 15.
2. In this case, the resistance remains constant. Let's represent the initial voltage, current, and resistance as V1, I1, and R1, respectively.
According to Ohm's Law, we have V1 = I1 * R1.
If the voltage decreases by a factor of 12, we can denote the new voltage as V2 = V1 / 12.
Substituting this into Ohm's Law, we get V2 = I1 * R1.
We can solve this equation for the new current, I2, by rearranging:
I2 = V2 / R1 = (V1 / 12) / R1 = V1 / (12 * R1)
So, when the voltage decreases by a factor of 12, the current will also decrease by the same factor of 12.
3. The device used to smooth out the pulsed DC to a smoother DC is called a capacitor. A capacitor works by storing the electrical energy from the pulsed DC and then releasing it gradually, resulting in a smoother and more constant DC voltage output. The capacitor acts as a temporary energy storage device, helping to reduce the fluctuations or ripples in the DC output and providing a more stable power supply for the electronic device.