Quizz #3 solutions
PROBLEM 1
Resistance is equal to R = ρL/A, where L is the length and A is the cross-sectional area. Applying this formula
ρL
to the problem, we get R = π(rρL
2
2 . [Due to the confusion of r2 being diameter or radius, answer R = π(r 2 /4−r 2 ) is
2 −r1 )
2
1
also accepted.]
PROBLEM 2
a. Capacitors C2 and C3 are in parallel, so can be replaced by a capacitor C23 = C2 + C3 . Then, capacitors
C1 and C23 are in series, so the total capacitance Ctot can be calculated via the formula C1tot = C11 + C123 , so
Ctot =
C1 ·C23
C1 +C23
=
C1 ·(C2 +C3 )
C1 +C2 +C3 .
b. Let’s combine capacitors 2 and 3 into C23 , so now C1 and C23 are in series. Charge conservation tells us that
1 (C2 +C3 )
1 C23
charge on C1 and C23 is the same, q1 . The total voltage is ε = q1 /C1 + q1 /C23 , so q1 = CεC1 +C
= εC
C1 +C2 +C3 .
23
Voltage on C2 is the same as the voltage on C3 and the same as the voltage on C23 and their charges q2 and q3 sum
q1 C3
εC1 C3
2
1
1 C2
= Cq22 = Cq33 , so q2 = qC1 C
= C1εC
up to q1 on C23 : q2 + q3 = q1 . Cq23
+C2 +C3 . Analogously, q3 = C23 = C1 +C2 +C3 .
23
c. Eenergy U =
2
1Q
2 C ,
so U1 =
q12
2C1
=
ε2 C1 (C2 +C3 )2
2(C1 +C2 +C3 )2 ,
U2 =
q22
2C2
=
ε2 C12 C2
2(C1 +C2 +C3 )2 ,
U3 =
q32
2C3
=
ε2 C12 C3
2(C1 +C2 +C3 )2 .
PROBLEM 3
a. Lightbulb is labeled with rms values, so Irms = Prms /Vrms = 100/115 A ≈ 0.87 A.
√
b. Amplitude of the current is 2 times more, so I = Imax sin (ωt + φ0 ), where φ0 is an arbitrary initial phase and
ω = 2πf , where f is the frequency. Most countries around
√ the world use 50 Hz or 60 Hz, in US the frequency is
set to be f = 60 Hz. Setting φ0 = 0 and as Imax = Irms 2, plugging in the numbers, we get I ≈ 1.23 sin (377t),
where time t is in seconds.
PROBLEM 4
a. Let’s call I a current going around this circuit in the clockwise direction. If I < 0, then actual current goes
in counter-clockwise direction. Then applying Kirchhoff’s voltage loop rule in clockwise direction gives us
1
−IR1 + ε2 − IR2 − ε1 = 0. So, I = Rε21 −ε
+R2 .
ε2 −ε1 2
2
1
b. Dissipated power in DC regime is equal to I 2 R, so in R1 it’s P1 = R1 ( Rε21 −ε
+R2 ) and in R2 it’s P2 = R2 ( R1 +R2 ) .
PROBLEM 5
R2
2
a. Equivalent resistance of the two resistors is R = RR11+R
, so the current is I = V /R = A cos (2πf t + φ0 ) RR11+R
R2 ,
2
where φ0 is the phase at t = 0. [Answers proportional to cos 2πf t and sin 2πf t are also accepted]
√
2
b. RMS current for sinusoidal waves is Imax / 2, so Irms = √A2 RR11+R
R2
c. To calculate rms power dissipated in R1 and R2 we need to find rms current flowing through R1 and R2 . Voltage
across R1 is the same as across the battery, so instantaneous current through R1 is I1 = V /R1 , so rms current
Vmax
A2
is I1rms = √
. RMS power in R1 is then P1rms = (I1rms )2 · R1 = 2R
. Calculation for the second resistor is
1
2R
1
analogous and results in P2rms =
A2
2R2 .
2
d. Total RMS power is Irms
· Rtotal , so Prms =
A2 R1 +R2
2 R1 R2 .
2
2
2
2
A
A
2
e. We want to check that total Prms = P1rms + P2rms . It’s true: A2 RR11+R
R2 = 2R1 + 2R2 . [We could also check that
2
instantaneous power produced by the source dissipates in resistors. Ptot = IV = A2 cos2 (2πf t + φ0 ) RR11+R
R2 .
2
A
P1 = V12 /R1 = R
cos2 (2πf t + φ0 ) and by analogy P2 = V22 /R2 =
1
A cos (2πf t + φ0 ). We can see that indeed Ptot = P1 + P2 .]
A2
R2
cos2 (2πf t + φ0 ). Here V1 = V2 = V =