MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Problem Solving Solutions 7: Building Simple Circuits
Open up Circuit Construction Kit (AC+DC), Virtual Lab
http://phet.colorado.edu/en/simulation/circuit-construction-kit-ac-virtual-lab
Question 1: Using voltage
b) The voltage across the battery and light bulb should be the same.
c) The brightness of the bulb changes with the voltage. Note that in real incandescent bulbs, the
filament which produces the light heats up considerably resulting in a drop in resistance and hence
the current (and thus the brightness) increases even faster with voltage.
Question 2: Measuring Current directly with the Ammeter
Adding a resistor in parallel with the bulb results in a smaller effective resistance so the overall
resistance drops and the current flowing from the battery will go up.
Question 3: Two bulbs in Series
Adding a second bulb in series increases the effective resistance so the current flowing from the
battery will go down. The same current flows through all circuit elements in series so the current
through the first bulb will go down and it will dim. Since the current through the first bulb goes down,
the voltage across it will also go down. Since both current and voltage go down in the first bulb, the
power goes down “doubly”. The battery voltage doesn’t change but its current goes down so its power
output goes down as well.
Question 4: Two Bulbs in Parallel
Adding a second bulb in parallel decreases the effective resistance so the current flowing from the
battery will go up. However, the voltage across the first bulb doesn’t change so the current through it
will stay the same and it will not brighten or dim. Since both current and voltage stay the same in the
first bulb, the power stays the same. The battery voltage doesn’t change but its current goes up so its
power output goes up as well.
Question 5: Concept Question
a) Bulb A does not turn on, because the voltage difference across it is zero.
b) Bulb B does not turn on, because the voltage difference across it is zero.
c) Bulb C does not turn on, because the current flowing through it is zero.
All three are true because in each case there is a wire which shorts out the selected bulb. The
voltage from one end of a wire to the other is always zero.
Question 6: Concept Question
Adding bulbs in series and/or parallel and recalling that the resistance of wires is zero, the effective
resistance of the circuits are: A) 3R (add in series) B) R (wire shorts out two bulbs)
C) R (only 1 bulb) D) R/3 (add in parallel) E) R (only 1 bulb)
The voltages are: A)  (only 1 battery) B)  (only 1 battery) C)  (two batteries in parallel put out
the same voltage as 1 battery) D)  (only 1 battery) E) 2(two batteries in series put out twice the
voltage as 1 battery)
Since power is V2/R, we get: A) 2/3R B) 2/R
So, we get A<B=C<D<E which is option (a).
C) 2/R
D) 32/R
E) (2) 2/R
Question 7: Computing the Resistance of a Light Bulb
IR =
VR
15W IB = IR VBATT = VR + VBULB RBULB =
VBULB
IB Power=IV
Question 8: Four Bulb Circuit
Bulbs A & D have the same current so have equal brightness. Bulbs B & C have each half the current
of A & D (assuming all have identical resistance) so B & C are equal in brightness to each other but
dimmer than A & D.
After the break, no current can flow through bulb C because one end is disconnected so bulb C goes
out. Now, the other 3 bulbs are in series and all will have the same current and the same brightness.
However, we want to know how that compares to the brightness before the break. Removing bulb C
changes the effective resistance of the circuit from 2.5R to 3R so the overall current goes down. This
means that bulb A & D (which carry all of the current in both cases) are dimmer than before. On the
other hand, bulb B used to carry only half of the current and now it carries all of the current. Since the
total current only goes down by a factor of (2.5/3), the total current in B goes up and bulb B gets
brighter.
Problem 2: Power in Resistor Networks
Consider the circuit shown in the figure with each the value of each resistor and battery indicated.
a) Simplify the network as much as possible by adding resistors in series or parallel. Draw your
new circuit diagram.
Answer: The two middle resistors are in parallel so we can replace them by an equivalent resistance
Req =
R2 R3
(4W)(4W)
=
= 2W .
R2 + R3 (4W) + (4W)
So we see that all the resistors have the same value. This will make the calculations easier.
b) On your new circuit diagram showing the following choices: currents in each branch, and
circulation directions for at least two loops?
Answer: We redraw the circuit diagram with the following choices: currents in each branch, and
circulation directions for the two loops as shown in the figure below.
c) Apply current in equals current out at as many junctions as necessary?
Answer: Applying current in equals current out (current junction rule) to either node yields
I1 = I + I 4 .
(1)
d) Write down the circuit loop law for each loop.
Answer: The sum of the potential differences around each closed loop is zero so the loop on the left
yields
e1 - I1R - IR = 0 .
(2)
e 2 - I 4 R + IR = 0 .
(3)
The loop on the right yields
e) Determine the current through each resistor.
Answer: We can solve Eq. Error! Reference source not found. for I1 in terms of I ,
e1
R
- I = I1
(4)
and solve Eq. Error! Reference source not found. for I 4 in terms of I ,
e2
R
+ I = I4 .
(5)
Now substitute Eqs. Error! Reference source not found. and Error! Reference source not found.
into Eq. Error! Reference source not found.
e1
R
-I =
e2
R
+ I + I.
(6)
and solve for I
I=
e1 - e 2
3R
=
50 V - 20 V
= 5 A.
6W
(7)
From Eq. Error! Reference source not found., the current passing through resistor with value
R1 = 2 W is
I1 =
e1
R
-I =
50 V
- 5 A=20 A
2W
(8)
From Eq. Error! Reference source not found.. the current passing through resistor with value
R4 = 2 W is
I4 =
e2
R
+I=
20 V
+ 5 A = 15 A .
2W
(9)
Because the two middle resistors are the same the current I = 5 A splits into two equal amounts and so
the currents I 2 and I 3 going through resistors with values R2 = 4 W and R3 = 4 W have the same
value and are equal to
I 2 = I 3 = 2.5 A .
f) Determine the power delivered to each resistor.
Answer: The power delivered to each resistor is calculated as
Pi = Ii e i = Ii ( Ii Ri ) = I i 2 Ri
Therefore,
PR1 = I R12 R1= (20A)2(2.0) = 800W
PR2 = I R2 2 R2 = (2.5A)2(4.0) = 25W
PR3 = I R32 R3 = (2.5A)2(4.0) = 25W
PR4 = I R4 2 R4 = (15A)2(2.0) = 450W
(10)