The answer to this depends on where the ammeter is in the circuit. Assuming the ammeter is in series with the bulb and no other objects are attached, then the current is 0.4 Amps. Otherwise, your question is unanswerable without more information.
The failed bulb breaks the circuit so no current can flow - so the other bulb goes out (but is OK).
Parallel circuit
A sourcing output supplies current or voltage to the load. The output would be hooked to one side of the load, and the other side is grounded. A sinking output works by pulling the voltage or current to ground. In this case, the load is connected to the 'hot' power supply terminal, and the other end is connected to the sinking output. When the output goes active ("goes low"), current flows from the power supply, through the load, then to ground via the output terminal. Sinking outputs are sometimes called 'open collector' outputs. You can visualize it this way: Take a light bulb and hook one lead to the chassis of your car. You just grounded the load. Now take a wire and hook it to the positive battery terminal. When you touch the battery wire to the bulb, it lights. You are 'sourcing' current to the bulb. Now, connect the bulb to the positive battery terminal. Take a wire and connect it to the car's frame (ground). The bulb already has a voltage supply - the positive battery terminal. When you touch the ground wire to the bulb, sinking current to ground, the bulb lights. The big deal here is IC's that sink current typically have much greater current capability than those with sourcing outputs. So why even bother with sourcing outputs? Their main claim to fame is the load can be ground-referenced, which is very desirable in some situations.
The forward current of an LED is current that goes from the anode of the LED to the cathode (the forward direction).
series
The failed bulb breaks the circuit so no current can flow - so the other bulb goes out (but is OK).
When a charged capacitor is connected to a light bulb, the current flows from the capacitor through the bulb, causing it to light up. Initially, the bulb may be very bright as the capacitor discharges quickly, but as time goes on, the brightness decreases as the capacitor loses its charge and the current flowing through the bulb decreases.
If one bulb in a series circuit goes out, then current can't flow anywherein the circuit. A circuit in which current can't flow is an open circuit.
SMOKE!!! Yiii-haaaa! An ammeter is always as low a resistance as possible. That way, the current that you measure in a circuit using the meter will be the same as if the meter were not there. So putting an ammeter in parallel with a circuit means that you just short circuited the circuit you were intending to measure. Poof! There goes your ammeter!An ammeter has a very low internal resistance. So, if it is connected in parallel with a load, it will short-out that load. The resulting high current flowing through the ammeter may severely damage the meter (and possible harm the user), although most are fitted with fuses to protect them.
You're connecting a resistance in parallel with the ammeter, so that when the meter and the additional resistor are inserted into a current-carrying circuit, a part of the current goes through the meter, and the rest of the current bypasses the meter and goes through the resistor instead. If you know exactly how much the split is, then you can calculate the larger total current from the smaller number that you read on the meter, so the meter is now able to measure and display more current than it could before ... the range of its measurement ability has been extended. Example: -- You know the resistance of the meter. -- You connect a resistor with exactly that same resistance in parallel with the meter. -- Now 1/2 of the current goes through the meter and 1/2 goes through the resistor. -- The total current is exactly double what you read on the meter.
Short answer, because it is designed for DC, not AC.In AC, the current goes back and forth all the time; the average current is zero. And that's what a galvanometer that is not especially designed for AC would show.
No. Circuit is open but still HOT.
The relationship between Volts, Amps and resistance is expressed by the following formulas. V = A*R, A = V/R, R = V/A, So to calculate the resistance of the lightbulb use R=220/10. Resistance equals 22 ohms. If you were to measure the resistance of the bulb with an ohm-meter you would get a lower reading. The resistance of the filement goes up as the bulb heats up. So the current (amps) are higher than 10 amps when the bulb is first turned on.
Making the important assumption that each bulb is equal in electrical resistance, the current will increase proportionally to the number of bulbs added (until the current limit of the battery is reached, that is). Clarification ... The current through the bulbs that were already there doesn't change, but the newcomer-bulbs add to the total current from the battery or power supply.
You don't. ...unless you want to directly measure the current in a circuit branch. That's the purpose of an ammeter. You can also use a volt meter if you know the resistance of a resistor in that branch to determine current (assuming DC circuit here) - current = voltage / resistance. This may be more useful for circuitry that is on a breadboard, since inserting an ammeter may not be practical.
bulb goes crazy
The thing that a bulb goes into is called a socket. It is the part that connects the bulb to the electrical supply in lighting fixtures.