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Original Correct Answer:
The voltage across an open switch should equal the supply voltage.
More Detailed Answer:
The above answer is basically correct. However, it may not be EXACTLY the supply voltage.
This is counter-intuitive at first glance and confuses a lot of people, including electrical engineers. The reason is this. Voltage, Current and Resistance are all interrelated by Ohms Law. Voltage equals Current multiplied by Resistance.
It is easy to think that since a switch is open, then you do not have current flow through the circuit. Thus, current times any Resistance is equal to zero volts because the current is equal to zero. Thus, by this logic, you would expect to find zero volts across the switch. In actuality this is true.
But, when you insert you meter, you change the conditions of the circuit, and the following is the result.
Let's say that you have a circuit with a resistance load like a heater, and a inductive load like a motor. The switch that powers these devices is open, thus their is no current flowing through the resistance or inductive loads.
Now you put the meter across the open switch. When you do, you insert a very large resistance in parallel to the open switch. Why? Because to get volts, the meter measures current flow through a known resistance, and then calculates voltage. To keep the resistance from impacting the circuit performance, the resistance is very large. Therefore, when you insert the meter, you will get a flow of current through the meter.
Because of this large resistance, the current trough the resistance load, inductive load, and wires is very small. Thus, the voltage drop across the loads and wiring is very small. Therefore, it appears that the entire voltage in the circuit is across the huge resistance in the meter. The result is a voltage reading that is very near the source voltage.
Let's do the math. Let's assume you have 120 volts. You also have a resistance of 500 ohms, and a motor winding that has 0 ohms resistance when DC is applied (This is true for motors). The meter has a 10 million ohm resistance.
If these loads are in series, the total resistance is 10Million 500 ohms. The 11.9 microamps. By multiplying the current flow to each resistance, you get 6 millivolts across the resistance, no voltage across the motor winding, and 119.994 across the open switch or meter. Since a meter rounds it reading, you would get 120.
If the loads are in parellel, you would get the same thing, becuase the switch is in series with both loads. In this example, the motor winding would have all the current flow through it since it is zero ohms, and the parallel resitance load is 500 ohms. Thus, the total resistance is the 10 million of the meter, and this resistance drive the current, and thus the largest voltage drop is at the swtich/meter. You could decide to remove the motor from this parellel circuit. If you did then the then the result is the series circuit above.
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∙ 11y ago
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