Download Electric Circuit Practice

Electric Circuit Practice
Three resistors are connected in series. If placed in a circuit with a 12-volt power
supply. Determine the equivalent resistance, the total circuit current, and the
voltage drop across and current at each resistor.
Three resistors are connected in parallel. If placed in a circuit with a 12-volt power
supply. Determine the equivalent resistance, the total circuit current, and the
voltage drop across and current in each resistor.
Combination Circuit
Analyze the following circuit and determine the values of the total resistance, total current, and the
current at and voltage drops across each individual resistor.
The analysis begins by using the resistance values for the individual resistors in order to
determine the equivalent resistance of the circuit.
Req = R1 + R2 + R3 = 11 + 7 + 20 = 38
Now that the equivalent resistance is known, the current through the battery can be
determined using Ohm's law equation. In using the Ohm's law equation ( V = I • R) to
determine the current in the circuit, it is important to use the battery voltage for V and
the equivalent resistance for R. The calculation is shown here:
Itot = Vbattery / Req = (12 V) / (38 ) = 0.31579 Amp
The 1.5 Amp value for current is the current at the battery location. For a series circuit
with no branching locations, the current is everywhere the same. The current at the
battery
2) The analysis begins by using the resistance values for the individual resistors in order
to determine the equivalent resistance of the circuit.
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 = (1 / 11 ) + (1 / 7 ) + (1 / 20 )
1 / Req = 0.283766
-1
Req = 1 / (0.283766
-1
)
Req = 3.52
(rounded from 3.524027 )
Now that the equivalent resistance is known, the current in the battery can be determined
using the Ohm's law equation. In using the Ohm's law equation ( V = I • R) to determine
the current in the circuit, it is important to use the battery voltage for V and the
equivalent resistance for R. The calculation is shown here:
Itot = Vbattery / Req = (12 V) / (3.524027 )
Itot = 3.41 Amp
(rounded from 3.4051948 Amp)
The 12 V battery voltage represents the gain in electric potential by a charge as it passes
through the battery. The charge loses this same amount of electric potential for any given
pass through the external circuit. That is, the voltage drop across each one of the three
resistors is the same as the voltage gained in the battery:
V battery = V 1 = V 2 = V 3 = 12 V
____________________________________________________________________
The first step is to simplify the circuit
The first step is to simplify the circuit by replacing the two parallel resistors with a single
resistor with an equivalent resistance. The equivalent resistance of a 4 and 6 resistor
placed in parallel can be determined using the usual formula for equivalent resistance of
parallel branches:
1 / Req = 1 / R1 + 1 / R2 + 1 / R3 ...
1 / Req = 1 / (3 ) + 1 / (6 )
1 / Req = 0.500
-1
Req = 1 / (0.500
-1
)
Req = 2.00
Based on this calculation, it can be said that the two branch resistors (R2 and R3) can be
replaced by a single resistor with a resistance of 2 . This 2 resistor is in series with R1
and R4. Thus, the total resistance is
Rtot = R1 + 2
+ R4 = 2
Rtot = 8
+2
+4
Now the Ohm's law equation ( V = I • R) can be used to determine the total current in
the circuit. In doing so, the total resistance and the total voltage (or battery voltage) will
have to be used.
Itot = Vtot / Rtot = (24 V) / (8
Itot = 3.0 Amp
)
The 3.0 Amp current calculation represents the current at the battery location. Yet,
resistors R1 and R4 are in series and the current in series-connected resistors is
everywhere the same. Thus,
Itot = I1 = I4 = 3.0 Amp
For parallel branches, the sum of the current in each individual branch is equal to the
current outside the branches. Thus, I2 + I3 must equal 3.0 Amp. There are an infinite
possibilities of I2 and I3 values which satisfy this equation. Determining the amount of
current in either branch will demand that we use the Ohm's law equation. But to use it,
the voltage drop across the branches must first be known. To determine the voltage drop
across the parallel branches, the voltage drop across the two series-connected resistors
(R1 and R4) must first be determined. The Ohm's law equation ( V = I • R) can be used
to determine the voltage drop across each resistor. These calculations are shown below.
V1 = I1 • R1 = (3.0 Amp) • (2
V4 = I4 • R4 = (3.0 Amp) • (4
) = 6.0 V
) = 12 V
This circuit is powered by a 24-volt source. Thus, the cumulative voltage drop of a charge
traversing a loop about the circuit is 24 volts. There will be a 18.0 V drop (6.0 V + 12.0
V) resulting from passage through the two series-connected resistors (R1 and R4). The
voltage drop across the branches must be 6.0 volts to make up the difference between
the 24 volt total and the 18.0 volt drop across R1 and R4. Knowing the voltage drop
across the parallel-connected resistors (R1 and R4) allows one to use the Ohm's law
equation ( V = I • R) to determine the current in the two branches.
I2 = V2 / R2 = (6.0 V) / (3
I3 = V3 / R3 = (6.0 V) / (6
) = 2.0 A
) = 1.0 A