Concept Questions with
Answers
8.02
W10D2
W10D2
DC Circuits
Today’s Reading Assignment W10D2 DC Circuits &
Kirchhoff’s Loop Rules Course Notes: Sections 7.1-7.5
2
Concept Q: Resistors In Parallel
Suppose that R1 >> R2 .
Then the equivalent
resistance
Req
R2
2. Req
R1
3. Req
R2 / 2
4. Req
R2 / 2
1.
3
Concept Q: Resistors In Parallel
Answer. When R1 >> R2
Then the equivalent
resistance is the smaller
resistance. This is an
important circuit design
principle.
1
1 1
= +
Req R1 R2
Req
1
R2
R2
4
Concept Question: Bulbs &
Batteries
An ideal battery is hooked to a light
bulb with wires. A second identical
light bulb is connected in parallel to
the first light bulb. After the second
light bulb is connected, the current
from the battery compared to when
only one bulb was connected.
1.
2.
3.
4.
Is Higher
Is Lower
Is The Same
Don’t know
P18- 5
Concept Question Answer: Bulbs &
Batteries
Answer: 1. More current flows from
the battery
There are several ways to see this:
(A) The equivalent resistance of the
two light bulbs in parallel is half that
of one of the bulbs, and since the
resistance is lower the current is
higher, for a given voltage.
(B) The battery must keep two
resistances at the same potential
 I doubles.
P18- 6
Concept Question: Bulbs &
Batteries
An ideal battery is hooked to a
light bulb with wires. A second
identical light bulb is connected in
series with the first light bulb. After
the second light bulb is connected,
the current from the battery
compared to when only one bulb
was connected.
1. Is Higher
2. Is Lower
3. Is The Same
P18- 7
Concept Question Answer: Bulbs &
Batteries
Answer: 2. Less current flows from the battery
The equivalent resistance of
the two light bulbs in series
is twice that of one of the
bulbs, and since the
resistance is higher the
current is lower, for the
given voltage.
P18- 8
Concept Question: Power
An ideal battery is hooked to a light bulb
with wires. A second identical light bulb
is connected in parallel to the first light
bulb. After the second light bulb is
connected, the power output from the
battery (compared to when only one
bulb was connected)
1.
2.
3.
4.
5.
Is four times higher
Is twice as high
Is the same
Is half as much
Is ¼ as much
P18- 9
Concept Question Answer: Power
Answer: 2. Is twice as high
The current from the
battery must double (it
must raise two light bulbs
to the same voltage
difference) and
P = I DV
P18- 10
Concept Question: Power
An ideal battery is hooked to a light
bulb with wires. A second identical
light bulb is connected in series with
the first light bulb. After the second
light bulb is connected, the light
(power) from the first bulb (compared
to when only one bulb was connected)
1.
2.
3.
4.
5.
Is four times higher
Is twice as high
Is the same
Is half as much
Is ¼ as much
P18- 11
Concept Question Answer: Power
Answer: 5. Is 1/4 as bright
R doubles  current is
cut in half. So power
delivered by the battery
is half what it was. But
that power is further
divided between two
bulbs now.
P=I R
2
P18- 12
Concept Question: Measuring
Current
If R1 > R2, compare the
currents measured by the
three ammeters:
1.
2.
3.
4.
5.
6.
7.
A1 > A 2 > A 3
A2 > A 1 > A 3
A3 > A 1 > A 2
A3 > A 2 > A 1
A3 > A 1 = A 2
None of the above
Not enough information is given.
P18- 13
Concept Question Answer:
Measuring Current
Answer: 4. A3 > A2 > A1
The total current must add to the
two individual currents, so A3
must be largest. Most current
prefers to go through the smaller
resistor so A2 > A1 .
P18- 14