The equation for the three values in the question will give the definite answer.
Amperage (I) is equal to the voltage (E) divided by the resistance (R).
I= E / R So as you can see the answer is True.
Example: 10 Volts and 50 Ohms in a circuit will have a current of .2 Amperes flowing through it. 10 / 50 = .2
You can also rearrange the equation to find the other two:
E= R * E
R= E / I
Voltage is equal to amperage time resistance. V=IR Therefore, I'd say voltage times amperage is equal to amperage squared times resistance. VI=IIR Really there's no point in multiplying the two. However, if you were to divide voltage by amperage, you would have the resistance of the circuit. V/I=R
Voltage and resistance determine amperage, assuming the source can provide the amperes.
This is a voltage drop question. The amperage of the circuit must be given. Without the load amperage this question can not be answered.
You can change current by altering potential difference or resistance... But assuming the voltage is constant, the resistance of the circuit restricts the flow of electrical current.
It will increase the current since the water heater is made of a heating element and which is resistive in nature. Ohms law states that V=IR where V is the voltage, I the current and R the resistance. Now the resistance will always remain constant. Thus, when the voltage is increased, the current will also increase.
If voltage remains constant and resistance is increased, the amperage will decrease per Ohm's Law.
A multimeter.
It is halved. coz voltage=current * resistance
Voltage will be constant. Resistance is dependent on the components in the circuit. Source: Electronics Technician for the US Govt
V = IR Where, V = voltage I = current R = resistance Thus if resistance is increased with constant voltage current will decrease
A multimeter.
Inversely. As resistance increases, current dereases; given that the applied voltage is constant.
Voltage is equal to amperage time resistance. V=IR Therefore, I'd say voltage times amperage is equal to amperage squared times resistance. VI=IIR Really there's no point in multiplying the two. However, if you were to divide voltage by amperage, you would have the resistance of the circuit. V/I=R
If resistance is halved while voltage remains constant, the current will double.
If the ratio of voltage to current is constant, then the circuit is obeying Ohm's Law. If the ratio changes for variations in voltage, then the circuit does not obey Ohm's Law.
No it cant. Voltage = Current x Resistance. So at constant Voltage if the Resistance is increased, Current will reduce
Here is the formula you use. I = E/R. I = amperage, E = volts, R = resistance in ohms.