Aim: To study open and short circuit conditions in a simple circuit SL No. Apparatus Specification Qty 1 Voltmeter 0 – 300V 3 2 Ammeter 0 – 2A 2 3 Batten Holder — 2 4 Incandescent lamp 40W, 230V 2 5 Switch SPST 2 Theory Imagine a dam. When the reservoir is full, water is […]

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]]>**Aim:** To study open and short circuit conditions in a simple circuit

SL No. | Apparatus | Specification | Qty |

1 | Voltmeter | 0 – 300V | 3 |

2 | Ammeter | 0 – 2A | 2 |

3 | Batten Holder | — | 2 |

4 | Incandescent lamp | 40W, 230V | 2 |

5 | Switch | SPST | 2 |

Imagine a dam. When the reservoir is full, water is supposed to fall off from the top, traveling the distance equal to the height of the dam. But you punch a hole at the bottom. Now the path of water is shortened, and the small hole you have made makes the whole dam collapse because of high pressure at the bottom. This is the principle of a short circuit.

In the same way, imagine a circuit in which current is supposed to travel from the positive terminal of a battery, into a wire which is connected to a bulb, and back to the negative terminal of the battery. But you shorten the path of current by directly connecting the positive terminal to the negative and the battery explodes. This is called short-circuiting.

In household, current has to travel through electrical appliances. But you are standing on the ground without any insulation between you and it, and you touch the live wire. Since you are at zero potential, (ground) the current (which is at higher potential) will travel through you, which is a short path for electricity And you will burst into flames. So you are causing a short circuit.

A short circuit is an electrical circuit that allows a current to travel along an unintended path with no or very low electrical impedance.

This results in an excessive current flowing through the circuit. The opposite of a short circuit is an “open circuit”, which is infinite resistance between two nodes

- Connect the circuit as shown in the circuit diagram
- Keep the switch 1 in off position and switch 2 in On position
- Note the potential difference across the two lamps, the main circuit current, the current through lamp L1 and supply voltage
- Repeat step 3 with both switches SW1 and SW2 On position
- Also, step 3 with both switches SW1 and SW2 Off position

**Open Circuit**: **Both** the switches are **Off**.**Closed Circuit**: Only **one** switch is **On**.**Short Circuit**: **Both** the switches are **On**.

Circuit condition | Supply Voltage V _{S} (Volts) | Total Current I (Amps) | Current Through lamp L _{1},I _{1} (Amps) | Voltage Across L _{1}, V_{1} | Voltage Across L _{2}, V_{2} |

Open | |||||

Closed | |||||

Short |

Potential difference and current during normal working conditions have been studied for simple circuit.

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]]>Aim: To measure electrical energy in a single-phase AC circuit using induction type energy meter. Apparatus Required SL No Apparatus Specification Qty 1. Energy Meter Single Phase300V, 10A 1 2. Wattmeter Single Phase300V, 5A, UPF 1 3. Variable Lamp Load Each lamp230V, 40W 1 4. Stopwatch 1 Theory The meter which is used for measuring the energy utilized by the electric load […]

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]]>**Aim**: To measure electrical energy in a single-phase AC circuit using induction type energy meter.

SL No | Apparatus | Specification | Qty |

1. | Energy Meter | Single Phase 300V, 10A | 1 |

2. | Wattmeter | Single Phase 300V, 5A, UPF | 1 |

3. | Variable Lamp Load | Each lamp 230V, 40W | 1 |

4. | Stopwatch | 1 |

The meter which is used for measuring the energy utilized by the electric load is known as the energy meter.

Single Phase induction type energy meter is also popularly known as ‘watt-hour meter’.

The energy meter has four main parts. They are the

- Driving System
- Moving System
- Braking System
- Registering System

The detail explanation of their parts is written below :

**1. Driving System** – The electromagnet is the main component of the driving system. It is the temporary magnet which is excited by the current flow through their coil. The core of the electromagnet is made up of silicon steel lamination. The driving system has two electromagnets. The upper one is called the shunt electromagnet, and the lower one is called series electromagnet.

The series electromagnet is excited by the load current flow through the current coil. The coil of the shunt electromagnet is directly connected with the supply and hence carries the current proportional to the shunt voltage. This coil is called the pressure coil.

The center limb of the magnet has the copper band. These bands are adjustable. The main function of the copper band is to align the flux produced by the shunt magnet in such a way that it is exactly perpendicular to the supplied voltage.

**2. Moving System** – The moving system is the aluminum disc mounted on the shaft of the alloy. The disc is placed in the air gap of the two electromagnets. The eddy current is induced in the disc because of the change of the magnetic field. This eddy current is cut by the magnetic flux. The interaction of the flux and the disc induces the deflecting torque.

When the devices consume power, the aluminum disc starts rotating, and after some number of rotations, the disc displays the unit used by the load. The number of rotations of the disc is counted at a particular interval of time. The disc measured the power consumption in kilowatt-hours.

**3. Braking system** – The permanent magnet is used for reducing the rotation of the aluminum disc. The aluminum disc induces the eddy current because of their rotation. The eddy current cut the magnetic flux of the permanent magnet and hence produces the braking torque.

This braking torque opposes the movement of the disc, thus reduces their speed. The permanent magnet is adjustable due to which the braking torque is also adjusted by shifting the magnet to the other radial position.

**4. Registration (Counting Mechanism)** – The main function of the registration or counting mechanism is to record the number of rotations of the aluminum disc. Their rotation is directly proportional to the energy consumed by the loads in the kilowatt-hour.

The rotation of the disc is transmitted to the pointers of the different dial for recording the different readings. The reading in kWh is obtained by multiply the number of rotations of the disc with the meter constant.

- Connect the circuit as shown in the circuit diagram.
- Note down the Energy-meter constant ‘K’ rotations/KW-Hr using the same determine the energy consumption recorded in KW-Hr per rotation of the Energy-meter disc.
- Switch on the supply. With all the lamps On, Note down the time for ‘n’ rotations of the energy meter disc. Also, note down the wattmeter reading.
- Determine the energy consumption of the lamp during ‘n’ rotations of the disc by both the energy-meter and wattmeter method using appropriate formulas as given.
- Repeat the above procedure for different number of disc rotations.

No of rotations ‘n’ | Time ‘t’ for ‘n’ rotations | Wattmeter Reading ‘W’ Watts | Energy Calculated E=W*t / (36*10^{5}) | Energy Indicated by the energy meter E=n/KKW-Hr |

4 | ||||

8 | ||||

12 |

Energy meter constant K = _______ Rotations/KW-Hr

Energy recorded per rotation of the disc, 1/K = ________KW-Hr

Wattmeter Constant, K_{1}= VIcos**ϕ**/ Full scale deflection = ___

- Energy Consumption as recorded y Energy meter when ______ lamps were on ______ KW-Hr
- Energy Consumption as calculated using wattmeter when _____ lamps were On _______ KW-Hr

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]]>Aim: To measure the total 3-phase power of 3-phase balanced load using 2-wattmeter method and verify the result with its theoretical value. Apparatus Required SL No. Apparatus Specification QTY 1 Voltmeter 0 – 600V 1 2 Ammeter 0 – 5A 1 3 Wattmeter 600V, 5A, UPF 2 4 3-Phase balanced resistive load 10KW 1 Theory […]

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]]>**Aim**: To measure the total 3-phase power of 3-phase balanced load using 2-wattmeter method and verify the result with its theoretical value.

SL No. | Apparatus | Specification | QTY |

1 | Voltmeter | 0 – 600V | 1 |

2 | Ammeter | 0 – 5A | 1 |

3 | Wattmeter | 600V, 5A, UPF | 2 |

4 | 3-Phase balanced resistive load | 10KW | 1 |

Two wattmeter method can be employed to measure the power in a 3 phase, three wire star and delta connected balanced load.

In two wattmeter method the current coils of the wattmeter are connected in series with two lines, say R and B and the potential coil of each wattmeter are joined to the third line i.e. Y as shown in the figure below.

- Connect the circuit as shown in the circuit diagram 1
- Switch on the 3-phase supply using TPST switch
- Note down the line voltage, line current and readings of the voltmeter
- Compare the total power observed by the two wattmeter method with its theoretical value

- Connect the circuit as shown in the circuit diagram 2
- Repeat steps 2 to 4 of the star connection procedure

3 Phase Connection | Line Voltage V_{L} | Line Current I_{L} | Power factor cosϕ | W_{1} | W_{2} | Total 3 phase power (Practical) P=W _{1} + W_{2} (watts) | Total 3 phase power (Theoretical) P= √ 3 V _{L}I_{L }cosϕ (watts) |

STAR | 1 | ||||||

STAR | 1 | ||||||

STAR | 1 | ||||||

DELTA | 1 | ||||||

DELTA | 1 | ||||||

DELTA | 1 |

Before writing wattmeter reading multiply with wattmeter constant :

wattmeter constant K = VI cos**ϕ** / full scale deflection

where ‘cos**ϕ**‘ is power factor of the wattmeter

Theoretically Calculated Power | Practically Measured Power |

Theoretically Calculated Power | Practically Measured Power |

By conducting the experiment we found out practically that two wattmeters are sufficient to measure the power of 3 phase load.

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]]>Aim: To practically determine the relation between line and phase quantities of both current and voltage and compare them with their respective theoretical values. Sl.No Apparatus Specification Qty 1. Voltmeter 0 – 600V 2 2. Ammeter 0 – 5A 2 3. 3 Phase Lamp Load ————– 1 In a single-phase system, there are two wires […]

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]]>**Aim**: To practically determine the relation between line and phase quantities of both current and voltage and compare them with their respective theoretical values.

Sl.No | Apparatus | Specification | Qty |

1. | Voltmeter | 0 – 600V | 2 |

2. | Ammeter | 0 – 5A | 2 |

3. | 3 Phase Lamp Load | ————– | 1 |

In a single-phase system, there are two wires which carry power from source to destination.

One among them is called ‘phase’ and the other is ‘Neutral’.

Neutral has a potential difference of 0 Volts.

The phase has a voltage that keeps on swinging between a positive and negative maximum with reference to neutral.

In a symmetric 3-phase supply system, three conductors carry alternating current of same frequency and voltage amplitude relative to a common reference but a phase difference of one-third of a cycle between each.

The common reference is usually connected to ground and often to a current-carrying conductor called the neutral.

Due to the phase difference, the voltage on any conductor reaches its peak at one-third of a cycle after one of the other conductors and one-third of a cycle before the remaining conductor. This phase delay gives constant power transfer to a balanced linear load.

As compared to a single-phase AC power supply that uses two conductors (phase and neutral), a three-phase supply with no neutral and the same phase to ground voltage with the same current capacity per phase can transmit three times as much power using just 1.5 times as many wires.

Line voltage refers to the voltage measured between any two line conductors in a balanced three-phase system.

Phase voltage refers to the voltage measured across any one component in a balanced three-phase source or load.

Wye-connected sources and loads always have line voltages greater than phase voltages, and line currents equal to phase currents.

Delta-connected sources and loads always have line currents greater than phase currents, and line voltage equal to phase voltage.

- connect the circuit as shown in figure 1
- Switch on the power supply using TPST(Triple pole single throw) switch
- Note down the line voltage, line current, phase voltage and phase current from the respectively connected meters

- Connect the circuit as shown in figure 2
- Repeat the steps 2 and 3 from the procedure above

Star Connection | line to line V_{L}(Volts) | line to neutral V_{ph} (Volts) | line I_{L} (Amps) | line I_{ph} (Amps) |

Practical | ||||

Theoretical | V_{ph} = V_{L} / √ 3 | I_{ph} = I_{L} |

Delta Connection | line to line V_{L}(Volts) | line to neutral V_{ph} (Volts) | line I_{L} (Amps) | line I_{ph} (Amps) |

Practical | ||||

Theoretical | V_{ph} = V_{L} | I_{ph}= I_{L} / √ 3 |

Star Connection | Delta Connection | |

Line voltage, V_{L} | ||

Phase Voltage, V_{ph} | ||

Line Current, I_{L} | ||

Phase Current, I_{ph} |

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