Download Ph 213 – Challenging Problems (set 6) Name: Due: August 6, 2013

Ph 213 – Challenging Problems (set 6)
Name: _______________________
Due: August 6, 2013
1. For the circuit shown at right, Let R1 = R3 =  and R2 = , and E1 = 2V, E2 = E3 = 4V, you
can neglect the internal resistance of all batteries.
A) Calculate the current through each battery.
B) Calculate the power delivered or
used (specify which case) by each
battery
A) Let’s choose the direction of
currents and travel direction
around the loop as shown on the
diagram.
Applying the junction rule for current
I1 = I2 + I3
Next apply loop rule for the two loops:
2 – 10I1 – 20I2 – 4 = 0  I1 = –2I2 – 0.2
4 + 20I2 – 10I3 + 4 = 0  I3 = 2I2 + 0.8
Insert in first equation: –2I2 – 0.2 = I2 +2I2 + 0.8
 I2 = – 0.2 A
(negative sign indicates the actual direction of current in R1 is to the left).
And I1 = –0.2A (negative sign indicates the actual direction of current in R1 is to the left).
And I3 = 0.4A
B) The power delivered by battery 1 is
The power delivered by battery 2 is
The power delivered by battery 3 is
P1 = E1 I1
P2 = E2 I2
P3 = E3 I3
 is
 is
 is
P1 = 2V . 0.2A = 0.4 w
P2 = 4V . 0.2A = 0.8 w
P3 = 4V . 0.4A = 1.6 w
2. In the circuit shown let R1 = R, R2 = 2R,
R3 = R, and R4 = 3R. The switch S has
been already in position 1 for a long
time. Do all calculations in terms of R,
C, E, and t (time)
A) What is the current (including
direction) through each of the four
resistors?
No current flows through resistor R3. A long time after the switch has
been closed, capacitor will be fully charged and acts as open in circuit,
so no current flows through the branch of the circuit containing the
capacitor and resistor R4. So the circuit looks like the circuit diagram at
right.
B) Find the potential VC across the capacitor.
The potential across the capacitor is the same as the potential across resistor 2. The circuit is
similar to the one below
VC = VR2 =
 R2 
 
VC  VR2  
 R1  R2 
2
VC  
3
C) Now let t = 0 when you move the switch S to position 2. Immediately after switch is moved to
position 2, what is the current (include direction) through each resistor?
When the switch is moved to position 2 the circuit looks
like the circuit diagram shown below. The current flows
counterclockwise (up from the capacitor). Because VC= (2 /
3) E and the equivalent resistance is Req= R4 + R3 = 4R, the
current is
I
VC
Req
2

3

4R

I

6R
D) Make a graph of current vs. time for the current that flows out of the capacitor after the switch
is moved to position 2 at t = 0. Indicate the value of the current at time t = 0, t = and t = 2 on
your graph.
Because the equivalence resistance is Req= 4R , the
time constant is τ= RC = 4RC. Recall in one time
constant τ the current drops to 1/e I0 = 0.37 I0
E) After keeping the switch S in position 2 for a long
time, the switch is thrown to position 1 again.
Immediately after the switch has been thrown to
position 1, find the current through the battery.
After a long period in position (2) the capacitor is now uncharged. An uncharged capacitor in a
closed circuit acts like a short. Therefore immediately after the switch has been thrown to
position (1), the capacitor can be replaced by a wire,
and the circuit now looks like one at right.
Resistors R2 and R4 are in parallel and result in series
with R1.
RR
I   /( R1  2 4 ) 
R2  R4
I   /( R 
6R 2
5
)
5R
11R