3 volts because an uncharged capacitor is equivalent to 0V source (treat it as a shorting wire), the circuit is not closed so there is no current, so no voltage drop on the resistor -- thus the entire 3V from the voltage source will be present between switch terminals
You will measure 3 volt. However, measuring long enough, the volt meters internal resistance will make a small current flowing. This current will charge the capacitor slowly. So that the volt meter will measure smaller and smaller voltage.
1. Assuming capacitor is not charged, the voltage over the switch is Vbat. 2. When the capacitor is charging, the bulb will be lit, and eventually dim. 3. When you disconnect the switch, it's going to measure near 0V. 4. Waiting for the internal resistance of the capacitor to drain, the probe will start measuring slowly up to Vbat again. This will likely take a loooooong time.
Great question. I'll say the short section with the light bulb will show ghost voltage on the switch end, 3V, but the number of electrons is so low, voltage instantly drops to zero when the switch is closed. With nowhere to go, the electrons trapped in the section of wire with the bulb shift away from the negatively charged plate to the switch end, but have nowhere to go. When switch is closed, electrons find the path back to the battery, voltage across the plate grows and the light briefly glows.
3 volts because an uncharged capacitor is equivalent to 0V source (treat it as a shorting wire), the circuit is not closed so there is no current, so no voltage drop on the resistor -- thus the entire 3V from the voltage source will be present between switch terminals
Starting it will be 3V if you leave the leads connected it will very slow increase as the capacitor changes
0V
The capacitor acts as an open. It’s zero volts dc
You will measure 3 volt. However, measuring long enough, the volt meters internal resistance will make a small current flowing. This current will charge the capacitor slowly. So that the volt meter will measure smaller and smaller voltage.
3volts
3v subtract 0
3volts.
1. Assuming capacitor is not charged, the voltage over the switch is Vbat.
2. When the capacitor is charging, the bulb will be lit, and eventually dim.
3. When you disconnect the switch, it's going to measure near 0V.
4. Waiting for the internal resistance of the capacitor to drain, the probe will start measuring slowly up to Vbat again. This will likely take a loooooong time.
Ghost voltage , no proper reference , this is what you get.
An uncharged capacitor in this case is an insulator then voltage across open switch is zero volt
Great question. I'll say the short section with the light bulb will show ghost voltage on the switch end, 3V, but the number of electrons is so low, voltage instantly drops to zero when the switch is closed. With nowhere to go, the electrons trapped in the section of wire with the bulb shift away from the negatively charged plate to the switch end, but have nowhere to go. When switch is closed, electrons find the path back to the battery, voltage across the plate grows and the light briefly glows.
My answer is 3 volts still, it's still a direct current from the battery to the switch
You will short circuit the circuit with the volt meter, so the voltage will rise very slowly.