Resistance of the load = voltage across the load/current through the load
Which means that the resistance would be 3 ohms.
(This is only true assuming that the load is purely resistive and the ammeter is ideal.)
The positive terminal of the battery would be connected to the positive terminal of the ammeter. The load would then be connected between the two negative terminals, positive side of the load being connected to the negative side of the ammeter.
the bulb will glow and ammeter will show the reading
The ammeter will only read when connected properly. If the meter is connected backwards, then the meter reads the wrong way. If it is a digital ammeter then it may not work at all.
Even though it is connected to a 9 volt source, it is still a resistor.
The Positive side of the Ammeter should be connected to the positive side of the battery and the Negative side should be connected to the Globe/Conductor/Insulator to provide a valid reading.
If a 9.0 volt battery is connected to a 4.0-ohm and 5.0-ohm resistor connected in series, the current in the circuit is 1.0 amperes. If a 9.0 volt battery is connected to a 4.0-ohm and 5.0-ohm resistor connected in parallel, the current in the circuit is 0.5 amperes.
a battery and a resistor connected together.
i=v/r can be used to it
All the way along when the crocodile clip is connected to a resistor, when the other end of the resistor is connected to the other side of the battery.
No. An ammeter is to be connected in series, between the device and the battery's positive output, Its NEGATIVE terminal (red lead) has to be connected to the POSITIVE of the battery. Its positive terminal (black lead) will then be connected to the device's positive terminal. (Connecting an ammeter in series with a power supply by itself may damage or destroy the meter.)Almost every single digital ammeter made will indicate reverse current, so the worst you might get is a displayed reading of a negative amperage if you connected it in reverse.(For an illustration, see the related link)
The ammeter is reading zero because there is no current flowing. This is because one of the resistors is faulty; the faulty resistor has an "open circuit" (open circuit means there is a broken connection). We know that: Ohms law is: V = I x R (voltage = current x resistance) Therefore because there is zero current in each resistor there will be zero voltage across each resistor. However we also know that: Kirchhoff's voltage law is: V1 +V2 +V3 + … = Vs (the sum of the voltage drops accross each component in a circuit MUST equal the supply (or battery) voltage). But if all the resistors are zero volts, then what component equals the supply (or battery) voltage? The battery voltage is developed across the open circuit… therefore the resistor which is faulty will have a voltage across it equal to the battery voltage. That easy to measure with a volt meter! hope this helps
The 12V battery connected to the 2.4 Ohm combination will supply 12/2.4 or 5A. The individual currents will be 12/3 or 4A for the 3 Ohm resistor, and 12/12 or 1A for the 12 Ohm resistor. The 2.4 Ohm parallel combination is obtained from a simple product-over-the-sum calculation.