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energy. CAPACITANCE
PRACTICE QUESTIONS
Question 1:
Solution:
The electric field is nonvanishing only in the region
obtain Using Gauss’ Law, we Therefore, the potential difference between the two conducting shells is: Answer 6.2
Capacitance increases when connected in parallel, decreases in series. To get the
maximum we close all 1 to 4 to put the three in parallel, and put 5 down to avoid putting
them in series.
To get the minimum we close 3 and 2, open 1 and 4, put 5 up, to put all four capacitors in
series to get the minimum. Solutions will be posted at www.prep101.com/solutions 22 ©Prep101 Phys1026 April Exam Solutions CURRENT AND RESISTANCE
PRACTICE QUESTIONS
Question 1:
Solution:
For A: V 2 (120 V )
P=
=
R
144 Ω 2 P = 100 W
For B:
The voltage drop across each bulb is 60 V, since they have equal resistances.
V 2 (60 V )
P=
=
R
144 Ω
P = 25 W 2 For C:
In the parallel configuration, the voltage drop across each bulb is 120 V, so the power
drawn by each bulb is 100 W  as in part a. Hence, the bulbs are much brighter.
Question 2:
Solution:
For A: dQ
90 C
=
dt 3600 s
i = 2.5 × 102 A i= For B:
The electron density in m3 is:
n = 5.8 × 10 22
j≡ electrons 10 6 cm 3
electrons
×
= 5.8 × 10 28
3
3
cm
1m
m3 i
= nqv d
A q = e = 1.60 × 1019 C where Note that since we are not worried about the direction of the current, we can ignore the
minus sign in the charge.
j≡ i
2.5 × 10 −2 A
A
=
= 3.18 × 10 4 2
2
−3
A π × 1.0×10 m
m
2 ( ) Solutions will be posted at www.prep101.com/solutions 23 ©Prep101 vd ≡ Phys1026 April Exam Solutions j
3.18 × 10 4 A ⋅ m −2
=
nq (5.8 × 10 28 m −3 )× (1.60 × 10 −19 C )
vd = 3.4 × 106 m/s 7.3 Answer Strategy draw a diagram of the situation
consider the internal resistance to be a single extra resistor in series
find the effective resistance of the resistors in parallel (Reff) The terminal voltage is half the emf – so fully half the voltage is dropped across the
internal resistance. Since the internal resistance and the rest of the circuit are in series,
this means that the internal resistance and the effective resistance of the rest of the circuit
are the same, ∴ Ri = Reff = 0.50Ω Solutions will be posted at www.prep101.com/solutions 24 ©Prep101 Phys1026 April Exam Solutions Answer 7.4 Strategy find R1 using Ohm’s Law
find V2, V3 using loop rule
IT splits into two equal parts
find R2 & R3 using Ohm’s Law again
First, all current goes through R1, so
R1 = V1 / IT
= (10.0V)/(10.0A)
= 1Ω
VT = V1 + V2
Find V3:
VT = V1 + V3
Find V2:
40.0V = 10.0V + V3
40.0V = 10.0V + V2
so V3 = 30.0V
so
V2 = 30.0V
Now,
I2 = I3 = ½ IT = 5.0A
Finally, combine the above to solve for resistances (by symmetry, R2 = R3)
R2 = V2/I2
= (30.0V)/(5.0A)
= 6.0 Ω
… = R3
*We could check this answer by finding Rtotal and applying it to R = V/I CIRCUITS
PRACTICE QUESTIONS
Question 1: Solution :
This is a question where Kirchhoff’s rules need to be used. There will be a fair amount of
substitution of equations into each other, so they will be numbered where necessary. Al...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.
 Winter '14
 Prof.PaulineBarmby
 Physics

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