In the given circuit diagram, calculate the magnitude of current L1 and l2 if radiation of ammetre is 4.2 A.
Answers
ANSWER
In the circuit shown, when key K is open branch BC will not be a part of the circuit (i.e. not connected to the circuit having battery ). The current will not flow through the branch BC having 4Ω resistor. Hence the ammeter will read the current flowing through remaining circuit.
Therefore,
I=
R
V
I=
5
10
I=2A
In the diagram above the ammeter is connected across the terminals of the 20v battery in parallel with the 10 ohm resistor. Most likely the ammeter will be at full deflection, as it will be functioning as a voltmeter.
Ammeters are used to measure current (measured in Amperes.) In order to measure a current we must measure the current THROUGH something. For purposes of experimentation and schematic drawings, we make the assumption that the interconnecting lines (representing wires or other connections) have no resistance. In real life though, all materials have resistive properties although some such as most metals have extremely low values.
Unfortunately the diagram doesn’t give us more information. For example, we don’t know the internal resistance of the ammeter. Internally an ammeter has a small resistance owing to the many turns of wire in the coil. There usually is also an internal (or external) “shunt” resistor. In my very first circuits class in college, one of our exercises was to calculate the internal resistance of a meter using known resistance.
If the ammeter was connected is SERIES (see diagram below) with the 10ohm resistor (instead of parallel) then we could use Ohm’s Law to figure out the current: I=E/R. I=current in Amperes, E=potential in Volts and R=resistance in Ohms.