For all practical purposes NO. The only voltage measured will be the voltage drop of the contact itself. This should be very low if the contact is good.
0.7 The voltage across a silicon diode when it is forward biased should be greater than or equal (>=) 0.7volts.
The voltage across a battery in a parallel circuit is equal to the voltage across each bulb because Kirchoff's Voltage Law (KVL) states that the signed sum of the voltages going around a series circuit adds up to zero. Each section of the parallel circuit, i.e. the battery and one bulb, constitutes a series circuit. By KVL, the voltage across the battery must be equal and opposite to the voltage across the bulb. Another way of thinking about this is to consider that the conductors joining the battery and bulbs effectively have zero ohms resistance. By Ohm's law, this means the voltage across the conductor is zero, which means the voltage across the bulb must be equal to the voltage across the battery and, of course, the same applies for all of the bulbs.
voltage across inductor create a flux. because of variation current developes an opposite emf.
Kirchoff's Voltage Law: The sum of the voltage drops across all elements in a series circuit add up to zero. If you know the voltage drops across all but one element, and you know the voltage rise across the source, then you can easily calculate the remaining drop.
In a series circuit the total voltage is the sum of the voltage drops across all the component in series. When the voltage drops across each the individual components are added up, they will equal the supply (or applied) voltage.
The voltage drop should be as close to zero as would be readable by a typical volt meter. If it is measurable you likely have a problem with corrosion or oxidation in switch that is increasing resistance. If you can measure a voltage drop across a closed switch contact, replace the switch. Or the switch is open, try flipping the switch!
The only way I can imagine getting these readings is if something is wrong with the circuit. If the circuit is off, then you get 0v on each of the readings, of course. In any case, if you are getting the readings as you describe, you have an electrical problem that will require a competent electrician to solve.
Since voltage is electrical potential difference, to measure the voltage across a component, you place the voltmeter across, or in parallel with, the component.
vary the rheostat by step by step note out the two voltmeter readings
0.7 The voltage across a silicon diode when it is forward biased should be greater than or equal (>=) 0.7volts.
what can be inferred about the voltage across the bulb
See the related link for a circuit diagram. The diagram pretty well explains how it works also. A potential is applied across the potentiometer. A movable contact moves across the voltage developed across the fixed portion and picks off the desired voltage.
How does the voltage measured across a dry cell ompare with the voltage drop measured across three bulbs in series?
Divide the number of lamps into the value of the supply voltage, and that will tell you the voltage drop appearing across each lamp. Bear in mind that if one of the lamps should be removed, the full supply voltage will then appear across the empty lamp holder.
The voltage measured across an open in a series circuit is the equivalent of the sourse voltage.
Without a circuit diagram, there are more questions before answers. First, assuming no other components, there is wire resistance to consider. There is also minor resistance in each connection, and each switch contact. In any event, the sum of the voltage drops in any series ckt equals the voltage of the source. If the sum of the wire, contact, and other voltage drops was .2 volts, and the source was 12.6 volts, then the light bulb would have 12.4 volts applied to it.
basically a diode flows an exponential curve Vs current if you try to double the voltage drop by increasing the voltage it should self destruct