Geoscience Reference
In-Depth Information
E
1
=
I
×
R
1
E
1
= 2.5 amps × 6 ohms
E
1
= 15 volts
Because
R
2
is the same ohmic value as
R
1
and carries the same current, the voltage drop across
R
2
is
also equal to 15 volts. Adding these two 15-volt drops together gives a total drop of 30 volts, exactly
equal to the applied voltage. For a series circuit then,
E
T
=
E
1
+
E
2
+
E
3
…
E
n
(11.27)
where
E
T
= Total voltage (V).
E
1
= Voltage across resistance
R
1
(V).
E
2
= Voltage across resistance
R
2
(V).
E
3
= Voltage across resistance
R
3
(V).
■
EXAMPLE 11.28
Problem:
A series circuit consists of three resistors having values of 10 ohms, 20 ohms, and 40
ohms, respectively. Find the applied voltage if the current through the 20-ohm resistor is 2.5 amp.
Solution:
To solve this problem, first draw a circuit diagram and label it as shown in Figure 11.31.
Given:
R
1
= 10 ohms
R
2
= 20 ohms
R
3
= 40 ohms
I
= 2.5 amps
Because the circuit involved is a series circuit, the same 2.5 amp of current flows through each
resistor. Using Ohm's law, the voltage drops across each of the three resistors can be calculated:
E
1
= 25 volts
E
2
= 50 volts
E
3
= 100 volts
R
1
10 ohms
+
R
2
20 ohms
E
= ?
-
40 ohms 2.5 a
R
3
FIGURE 11.31
Solving for applied voltage in a series circuit.
Search WWH ::
Custom Search