BASIC ELECTRONICS - OD1633 - LESSON 1/TASK 2
d. Source Resistance. A meter connected across the terminals of a good
1.5 volt battery reads about 1.5 volts. When the same battery is inserted
into a complete circuit, the meter reading decreases to something less than
This difference in terminal voltage is caused by the internal
resistance of the battery (the opposition to current offered by the
electrolyte in the battery). All sources of electromotive force have some
form of internal resistance which causes a drop in terminal voltage as
current flows through the source.
This principle is illustrated in figure 40 (on the following page) where the
In the schematic, the
internal resistance is indicated by an additional resistor in series with
the battery. The battery, with its internal resistance, is enclosed within
With the switch open, the
voltage across the battery terminals reads 15 volts.
When the switch is
closed, current flows causing voltage drops around the circuit. The circuit
current of 2 amperes causes a voltage drop of 2 volts across Ri. The 1 Ohm
internal battery resistance thereby drops the battery terminal voltage to 13
attempt to do so would damage the meter.
The effect of the source resistance on the power output of a dc source may
be shown by an analysis of the circuit in figure 41 on page 69. When the
variable load resistor (RL) is set at the zero Ohm position (equivalent to a
short circuit) current (T) is calculated using the following formula:
This is the maximum current that may be drawn from the source. The terminal
voltage across the short circuit is zero volts and all the voltage is across
the resistance within the source.
If the load resistance (RL) were increased (the internal resistance
remaining the same), the current drawn from the source would decrease. At
the same time, the terminal voltage applied across the load would increase
and approach a maximum as the current approaches zero amps.