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Removing any bulb breaks the continuity of a series circuit, stopping the flow of electrical current. Removing a bulb in a parallel circuit does not interrupt the current flow, so the remaining lights continue to conduct electrical current.
a meter uses its own supply to measure resistance. you dont have to remove the resistor from the circuit but you must isolate it from the circuit supply to get an accurate reading.
All other light bulbs will turn off due to the series circuit being broken. The electrons cannot travel all the way through, hence the current will also not flow in the circuit, switching off all the other bulbs too.CommentIt's worth pointing out that the full supply voltage will then appear across the lamp holder -so take care!
For carrying Short Circuit Test on Power Transformer Do the following: 1] Isolate the Power Transformer from service. 2] Remove HV/LV Jumps and Disconnect Neutral from Earth/Ground. 3] Short LV Phases by Cu/Al plate which could withstand short circuit current and connect these short circuited terminals to Neutral 4] Energise HV side by LV supply (440 3ph Supply) with OLTC tap position on Normal. 5] Measure Current in Neutral, LV line voltages, HV Volatage and HV Line Currents on various OLTC Tap position. Analysis: If Neutral current is near to zero transformer windings are OK If Neutral current is higher or equal to Line current between LV Phase one of the winding is Open.
Opening a circuit breaker merely interrupts the current flowing through that circuit. It doesn't remove the voltage from the circuit breaker. Isolators (called 'disconnects' in North America) place a visible gap between the circuit breaker and energised conductors, allowing work to be performed on the circuit breaker safely. The full procedure involves: 1. Tripping the circuit breaker (CB). 2. Opening both isolators. 3. Applying temporary earth (ground) connections on either side of the CB. 4. Placing barriers and warning signs around the CB. 5. Issuing a permit to work card to the maintenance supervisor.
The battery is the power source of the circuit. It supplies current to the circuit and the circuit is simply a path for the current to follow. When you remove the current (battery), the path still exists but there is no current going through it.
Increase, decrease, or remove the load <<>> Change the voltage and the current will also change in direct proportion, Ohms law.
Over load in circuit? Remove starter and have it checked at auto parts for excessive current draw Dead short to ground in circuit? Over load in circuit? Remove starter and have it checked at auto parts for excessive current draw Dead short to ground in circuit?
In a series circuit, all the current passes through the one circuit. Any break will totally remove power from all of the circuit.Parallel circuits have more than one branch where the current can flow. A broken wire will only affect one part, the rest of the circuit will still pass current.In a series circuit, all the current passes through the one circuit. Any break will totally remove power from all of the circuit.Parallel circuits have more than one branch where the current can flow. A broken wire will only affect one part, the rest of the circuit will still pass current.
All the bulbs will go out. In a series circuit, the current at all points is the same. This is Kirchoff's Current Law. If you loosen or remove a bulb in a series circuit, the current at that bulb becomes zero, and by Kirchoff's Current Law, the current in every part of the circuit also becomes zero.
Removing any bulb breaks the continuity of a series circuit, stopping the flow of electrical current. Removing a bulb in a parallel circuit does not interrupt the current flow, so the remaining lights continue to conduct electrical current.
Answer: it will burn out **Explain:**The same current flows through each part of a series circuit. If the circuit is broken at any point there won't be any current that will flow. In this case, if one of the bulbs blew out, the other bulb would not be able to light up because the flow of electric current would have been interrupted. #Carryonlearning
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.
if we remove a resistor from the parallel connection the effective resistance value will be increased.
First,remove all current and voltage sources ie replace voltage source with a short and keep current source open.Now draw the equivalent resistance-only circuit and find the equivalent resistance as viewed from the terminals of the circuit.
Overloading of electric circuit means that the current circulating in the circuit becomes more than the capacity of components in the circuit to withstand the current. All components in the circuits have some resistance passage of electricity through this resistance produces heat which is directly proportional to the square of current flowing. The components are designed to withstand only that much of heat as is generated by maximum designed current. When the current is more than this level, or in other words when the circuit is overloaded the components get overheated, leading to their damage. Frequently electric circuits in homes, factories, and other electrical installations incorporate fuses, which are essentially circuit component which protect other components in the circuit by quickly melting or burning out when the circuit is overloaded, resulting in breaking of the circuit. In common language this is called blowing of fuse.
You have to replace the wire (as you are increasing the current capacity), the outlet, and the breaker. Essentially you have to remove the old circuit and put in a new one. You can't reuse parts of the old circuit as you are increasing the current capacity and they would be underrated.