A galvanometer can be converted into ammeter by connecting a low resistance called shunt in parallel to the galvanometer. A voltmeter is a device used to measure the potential difference between two points in a circuit.

A galvanometer can be converted in to a voltmeter by connecting a high resistance in series connection within it. The scale is calibrated in volt. The value of the resistance connected in series decides the range of the voltmeter. The resistance is calculated by this equation which is connected in series.

Since galvanometer is a very sensitive instrument that cannot measure heavy currents in order to convert a galvanometer into ammeter we have to put a low resistance called shunt in parallel to it so that most of the current goes through it , and our device can measure the accurate value.

A galvanometer can be converted into a voltmeter by connecting a large resistance in series to the galvanometer which is shown in diagram. Let G and R be the resistance of a galvanometer and a conductor connected in series with it respectively.

To convert a moving coil galvanometer to a voltmeter, we add a high series resistance, but why?. The high resistance causes most of the voltage to drop across it, leaving a small voltage drop across the galvanometer. This also ends up making the voltmeter have very high resistance.

In order for an ammeter to measure a device’s current, it must be connected in series to that device. This is necessary because objects in series experience the same current. Ammeter in Series: An ammeter (A) is placed in series to measure current. All of the current in this circuit flows through the meter.

When ammeter is connected in parallel to the circuit, net resistance of the circuit decreases. Hence more current is drawn from the battery, which damages the ammeter.

1 ) Resistance of an ammeter is low. As ammeter is connected in series in the circuit, effective resistance is the sum of resistance of the circuit and the resistance of ammeter. Thus small resistance of ammeter would not effect the current in the circuit. 2) Resistance of Voltmeter is High.

The current is the ratio of the potential difference and the resistance. It is represented as (I). The current formula is given as I = V/R. The SI unit of current is Ampere (Amp).

amperes

Answer. Answer: the ammeter readings would be where A1 shows 1 ampere, A2 shows zero as bulb B2 blows off, A3 shows 1 ampere and A shows 2 amperes, the total current in the circuit.

position 1

Now potential difference across 3 ohm resistor, V = IR. so Potential difference across 3 ohm resistor V = 3*1/3 = 1 Volt.

The current can be found from Ohm’s Law, V = IR. The V is the battery voltage, so if R can be determined then the current can be calculated.

In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component.

The formula for calculating wattage is: W (joules per second) = V (joules per coulomb) x A (coulombs per second) where W is watts, V is volts, and A is amperes of current. In practical terms, wattage is the power produced or used per second. For example, a 60-watt light bulb uses 60 joules per second.

Voltage drop of the circuit conductors can be determined by multiplying the current of the circuit by the total resistance of the circuit conductors: VD = I x R.

The NEC recommends that the maximum combined voltage drop for both the feeder and branch circuit shouldn’t exceed 5%, and the maximum on the feeder or branch circuit shouldn’t exceed 3% (Fig. 1). This recommendation is a performance issue, not a safety issue.

Wire Sizing Chart and Formula

1. Calculate the Voltage Drop Index (VDI) using the following formula:
2. VDI = AMPS x FEET ÷ (% VOLT DROP x VOLTAGE)
3. Determine the appropriate wire size from the chart below.

4) in the National Electrical Code states that a voltage drop of 5% at the furthest receptacle in a branch wiring circuit is acceptable for normal efficiency. In a 120 volt 15 ampere circuit, this means that there should be no more than a 6 volt drop (114 volts) at the furthest outlet when the circuit is fully loaded.

The acceptable maximum voltage drop for DC loads is 5% of nominal battery voltage.

The simplest way to reduce voltage drop is to increase the diameter of the conductor between the source and the load, which lowers the overall resistance. In power distribution systems, a given amount of power can be transmitted with less voltage drop if a higher voltage is used.

As an example, for a 120-volt circuit, you can run up to 50 feet of 14 AWG cable without exceeding 3 percent voltage drop….For 120-volt circuits:

In general, bury metal conduits at least 6 inches below the soil surface. You may also run them at a depth of 4 inches under a 4-inch concrete slab. Under your driveway, the conduits must be below a depth of 18 inches, and under a public road or alleyway, they must be buried below 24 inches.

If the circuit is 100 amp or less you have to size the conductors based on the 60-degree celsius column unless the breaker and the equipment terminations are rated for 75 or 90 degrees. You have to use a #4 conductor to feed a 60 amp circuit.

A 12 gauge wire is typically good for 15 amp at 100 feet. For your 160 feet, they suggest a 6 gauge.

10 outlets

For general use receptacles, In commercial buildings it is limited to 180VA per duplex or single receptacle, therefore on a 12/2 Romex cable not otherwise subject to ampacity derating and protected by a 20 amp circuit breaker, that would allow a maximum of 13 receptacles.

Yes you can mix 12 and 14 gauge wire with a 15 amp breaker on the circuit. But it is not recommended because the next homeowner or electrician may be confused and put a 20 amp breaker on it again.