#5
Connect the Battery
Which is the correct way to
D) all are correct
light the lightbulb with the
battery?
A)
B)
C)
Connect the Battery
Which is the correct way to
D) all are correct
light the lightbulb with the
battery?
A)
B)
C)
Current can only flow if there is a continuous connection from the
negative terminal through the bulb to the positive terminal. This is
only the case for Fig. (3).
#6
Ohm’s Law
across a certain conductor
A) Ohm’s law is obeyed since the current
still increases when V increases
and you observe the current
B) Ohm’s law is not obeyed
You double the voltage
increases three times. What
can you conclude?
Ohm’s Law
across a certain conductor
A) Ohm’s law is obeyed since the current
still increases when V increases
and you observe the current
B) Ohm’s law is not obeyed
You double the voltage
increases three times. What
can you conclude?
Ohm’s law, V = I R, states that the
relationship between voltage and current is
linear. Thus for a conductor that obeys
Ohm’s Law, the current must double when
you double the voltage.
Follow-up: Where could this situation occur?
#10 Series Resistors I
A) 12 V
Assume that the voltage of the battery
is 9 V and that the three resistors are
identical. What is the potential
difference across each resistor?
B) zero
C) 3 V
D) 4 V
9V
Series Resistors I
A) 12 V
Assume that the voltage of the battery
is 9 V and that the three resistors are
identical. What is the potential
difference across each resistor?
B) zero
C) 3 V
D) 4 V
Since the resistors are all equal,
the voltage will drop evenly
across the 3 resistors, with 1/3 of
9 V across each one. So we get a
3 V drop across each.
9V
Follow-up: What would be the potential difference if
R= 1 W, 2 W, 3 W
#11 Series Resistors II
A) 12 V
In the circuit below, what is the
B) 6 V
voltage across R1?
C) 8 V
D) 4 V
R1= 4 W
R2= 2 W
12 V
Series Resistors II
A) 12 V
In the circuit below, what is the
B) 6 V
voltage across R1?
C) 8 V
D) 4 V
The voltage drop across R1 has
to be twice as big as the drop
across R2. This means that V1 =
R1= 4 W
R2= 2 W
8 V and V2 = 4 V. Or else you
could find the current I = V/R =
(12 V)/(6 W) = 2 A, then use
12 V
Ohm’s Law to get voltages.
Follow-up: What happens if the voltage is doubled?
#12 Parallel Resistors I
A) 10 A
In the circuit below, what is the
B) 5 A
current through R1?
C) 2 A
D) 7 A
R2= 2 W
R1= 5 W
10 V
Parallel Resistors I
A) 10 A
In the circuit below, what is the
B) 5 A
current through R1?
C) 2 A
D) 7 A
The voltage is the same (10 V) across each
R2= 2 W
resistor because they are in parallel. Thus,
we can use Ohm’s Law, V1 = I1 R1 to find the
R1= 5 W
current I1 = 2 A.
10 V
Follow-up: What is the total current through the battery?
#13 Short Circuit
Current flows through a
A) all the current continues to flow
through the bulb
connected across the
B) half the current flows through the wire,
the other half continues through the
bulb
bulb, what happens?
C) all the current flows through the wire
lightbulb. If a wire is now
D) none of the above
Short Circuit
Current flows through a
A) all the current continues to flow
through the bulb
connected across the
B) half the current flows through the wire,
the other half continues through the
bulb
bulb, what happens?
C) all the current flows through the wire
lightbulb. If a wire is now
D) none of the above
The current divides based on the
ratio of the resistances. If one of the
resistances is zero, then ALL of the
current will flow through that path.
Follow-up: Doesn’t the wire have SOME resistance?
#14 Circuits I
The lightbulbs in the circuit below
A) circuit 1
are identical with the same
B) circuit 2
resistance R. Which circuit
produces more light? (brightness
 power)
C) both the same
D) it depends on R
Circuits I
The lightbulbs in the circuit below
A) circuit 1
are identical with the same
B) circuit 2
resistance R. Which circuit
produces more light? (brightness
 power)
In #1, the bulbs are in parallel,
lowering the total resistance of the
circuit. Thus, circuit #1 will draw
a higher current, which leads to
more light, because P = I V.
C) both the same
D) it depends on R