If the three light bulbs are in parallel across the battery, then the full 9 volts appears
across each light bulb.
If the three light bulbs are in series across the battery, then the voltage divides among
them as follows:
Across the 10-ohm: 9 x (10/60) = 1.5 volts
Across the 20-ohm: 9 x (20/60) = 3 volts
Across the 30-ohm: 9 x (30/60) = 4.5 volts
Sum of the voltages across the 3 individual light bulbs = 1.5 + 3 + 4.5 = 9 volts.
When a parallel plate capacitor is connected to a battery, the voltage across the capacitor increases as it charges. The battery provides a potential difference that causes charges to accumulate on the plates, leading to an increase in voltage until the capacitor is fully charged.
You can measure the voltage of a light bulb by using a multimeter set to the voltage setting. Simply touch the positive and negative probes of the multimeter to the base of the light bulb where it connects to the socket to measure the voltage across the bulb.
Simply add all of the component's resistances together and that will give you circuits total resistance. If you're dealing with a 'series-parallel', or 'parallel' circuit, the equations will change, but in a simple series circuit, the total resistance is just the total of all the component's resistance.
The energy delivered by a battery would depend on-- the battery's voltage-- the resistance of the load connected across its output terminals-- the length of the time the load is connectedThe power delivered by the battery is [ (voltage)2 divided by (load resistance) ].The total energy delivered by the battery is [ (power) multiplied by (time the load is connected) ].
The rule for voltage across each resistor in a series circuit is that the total voltage supplied by the source is equal to the sum of the voltage drops across each resistor. In a parallel circuit, the voltage across each resistor is the same and equal to the source voltage.
The voltage across the resistors will remain constant.
forcing a constant current and measuring the voltage across the unknown resistor.
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.
There is only one possible answer.Voltage = current * resistancealso, Power = Voltage * Voltage / resistanceIf you follow the algebra, you should get a resulting resistance of the lightbulb to be 548.6 ohms.
In a parallel circuit, Voltage is constant through out the circuit. Thus, the voltage across each lamp is 6-volts.
The voltage drop across each resistor is determined by the amounts of resistance in the 3 resistors and all the rest of the resistances in the electrical circuit.
Voltage is energy per charge, in joules per coulomb, commonly known as the volt. It is produced by batteries, generators, current sources across resistances, voltage sources, thermocouples, solar cells, etc.
All the volt drops across each component in the series circuit will add up to the emf of the battery or power supply. Note the voltage supplying the circuit and divide it by the sum of all the resistances in the circuit. This gives you the current which is the same through all components. Take a component's resistance and multiply by the current and you have the voltage drop across it.
When a parallel plate capacitor is connected to a battery, the voltage across the capacitor increases as it charges. The battery provides a potential difference that causes charges to accumulate on the plates, leading to an increase in voltage until the capacitor is fully charged.
As the source impedance is equal to load impedance, it is assumed that both the resistances are in series. This distributes the whole voltage equally between both the impedance. Hence the PD across external resistance will be 1V.
(A) The bias battery voltage (B) 0V (C) the diode barrier potentiaol (D) The total circuit voltage
Both resistors will have the voltage of the battery.