False. Voltage (E) is the potential difference, i.e. electrical potential, in joules per coulomb. Current (I), on the other hand, is electrical charge flow, in coulombs per second. The two units are not related except through a common term such as resistance (R).
E = IR
I = E/R
R = E/I
False. Current is voltage divided by resistance.
false
Aim of any measuring instrument is to measure the object without affecting it. Voltmeter is used to measure voltage between two points and connected in parallel. Thus voltmeter should not change the voltage. If voltmeter resistance is very high, it will be as good as infinity compared to load. Thus connecting voltmeter will not change the voltage and measure it accurately.
Inductive. Used to remember this by "Eli" the "ice" man. "(e) Voltage (l) (Inductive circuit) (i) current", the ,"(i) Current (c) (capacitive circuit) (e) voltage, man.
You connect a galvanometer in series with the circuit being measured, because the galvanometer is a form of ammeter, although an extremely sensitive one, and ammeters measure the current in a series circuit.
Current will flow as long as there is a difference of potential (a voltage) and a path for current to flow. So no power-control device is required for current flow but yes it will flow with a power control decive.
False
True. Ohm's law states the voltage is resistance times current.
true
It's true, but only for a constant applied voltage. It turns out that the current in that wire is also proportional to the voltage applied to the circuit. Here's how it works.For a given applied voltage, current is (inversely) proportional to resistance. More resistance (for that same applied voltage) makes for less current will flow. Less resistance (for that same applied voltage) means more current will flow. We say that current and resistance are proportional because for a given voltage, when one goes up by "X" amount, the other goes down by the same proportion. Double the resistance, current is cut in half. Ten times the resistance will allow only one tenth the current. And we say inversely, because the change is in the "other" direction. When current goes up, resistance must have gone down (for that same applied voltage). Here's the other half of the idea with current and voltage and a given value of resistance.In a circuit with a fixed resistance, more voltage will drive more current. Less voltage will drive less current. They're directly proportional. That means if you double the voltage, you'll get double the current (through the same resistor). If you cut the voltage by a factor or ten (one tenth the applied voltage), you'll end up reducing current by a factor of ten (one tenth the current) for that same resistor. We usually think of circuits this way, but we can also think of them in other ways according to Ohm's law. Let's summarize.Voltage (E) is expressed in volts (V), and current (I) is expressed in amps (A), and resistance (R) is expressed in ohms (upper case omega).E = I x R (E equals I times R)R = E / I (R equals E divided by I)I = E / R (I equals E divided by R)AnswerFalse. Ohm's Law, in essence, states that the ratio of voltage to current is constant for variations in voltage. No mention of resistance whatsoever! However, for those materials that obey Ohm's Law ('linear' or 'ohmic'), raising a conductor's resistance will cause the current to fall (assuming the voltage is fixed). so the current must be INVERSELY proportional (not proportional) to the resistance!
false
Aim of any measuring instrument is to measure the object without affecting it. Voltmeter is used to measure voltage between two points and connected in parallel. Thus voltmeter should not change the voltage. If voltmeter resistance is very high, it will be as good as infinity compared to load. Thus connecting voltmeter will not change the voltage and measure it accurately.
Inductive. Used to remember this by "Eli" the "ice" man. "(e) Voltage (l) (Inductive circuit) (i) current", the ,"(i) Current (c) (capacitive circuit) (e) voltage, man.
If it is a true or false question your answer is: False. :)
false
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 = .2You can also rearrange the equation to find the other two:E= R * ER= E / I
True: Resistance R = voltage V / amperage IResistance is the conducting material.
That is false.
You connect a galvanometer in series with the circuit being measured, because the galvanometer is a form of ammeter, although an extremely sensitive one, and ammeters measure the current in a series circuit.