Electric Circuits Free Response Practice Questions | AP Physics C Electricity and Magnetism - Grade 9 PDF Download

 Page 1


Section II: Free Response
1. Consider the following circuit:
(a) At what rate does the battery deliver energy to the circuit?
(b) Find the current through the 20 O resistor.
(c) (i) Determine the potential difference between points a and b.
(ii) At which of these two points is the potential higher?
(d) Find the energy dissipated by the 100 O resistor in 10 s.
(e) Given that the 100 O resistor is a solid cylinder that’s 4 cm long, composed of a material
whose resistivity is 0.45 O m, determine its radius.
2. The diagram below shows an uncharged capacitor, two resistors, and a battery whose emf is e.
The switch S is turned to point a at time t = 0.
(Express all answers in terms of C, r, R, e, and fundamental constants.)
(a) Determine the current through r at time t = 0.
(b) Compute the time required for the charge on the capacitor to reach one-half its final value.
Page 2


Section II: Free Response
1. Consider the following circuit:
(a) At what rate does the battery deliver energy to the circuit?
(b) Find the current through the 20 O resistor.
(c) (i) Determine the potential difference between points a and b.
(ii) At which of these two points is the potential higher?
(d) Find the energy dissipated by the 100 O resistor in 10 s.
(e) Given that the 100 O resistor is a solid cylinder that’s 4 cm long, composed of a material
whose resistivity is 0.45 O m, determine its radius.
2. The diagram below shows an uncharged capacitor, two resistors, and a battery whose emf is e.
The switch S is turned to point a at time t = 0.
(Express all answers in terms of C, r, R, e, and fundamental constants.)
(a) Determine the current through r at time t = 0.
(b) Compute the time required for the charge on the capacitor to reach one-half its final value.
(c) When the capacitor is fully charged, which plate is positively charged?
(d) Determine the electrical potential energy stored in the capacitor when the current
through r is zero.
When the current through r is zero, the switch S is then moved to Point b; for the following
parts, consider this event time t = 0.
(e) Determine the current through R as a function of time.
(f) Find the power dissipated in R as a function of time.
(g) Determine the total amount of energy dissipated as heat by R.
Page 3


Section II: Free Response
1. Consider the following circuit:
(a) At what rate does the battery deliver energy to the circuit?
(b) Find the current through the 20 O resistor.
(c) (i) Determine the potential difference between points a and b.
(ii) At which of these two points is the potential higher?
(d) Find the energy dissipated by the 100 O resistor in 10 s.
(e) Given that the 100 O resistor is a solid cylinder that’s 4 cm long, composed of a material
whose resistivity is 0.45 O m, determine its radius.
2. The diagram below shows an uncharged capacitor, two resistors, and a battery whose emf is e.
The switch S is turned to point a at time t = 0.
(Express all answers in terms of C, r, R, e, and fundamental constants.)
(a) Determine the current through r at time t = 0.
(b) Compute the time required for the charge on the capacitor to reach one-half its final value.
(c) When the capacitor is fully charged, which plate is positively charged?
(d) Determine the electrical potential energy stored in the capacitor when the current
through r is zero.
When the current through r is zero, the switch S is then moved to Point b; for the following
parts, consider this event time t = 0.
(e) Determine the current through R as a function of time.
(f) Find the power dissipated in R as a function of time.
(g) Determine the total amount of energy dissipated as heat by R.
Section II: Free Response
1. (a) The two parallel branches, the one containing the 40 O resistor and the other a total of
120 O, is equivalent to a single 30 O resistance. This 30 O resistance is in series with the
three 10 O resistors, giving an overall equivalent circuit resistance of 10 + 10 + 30 + 10 =
60 O. Therefore, the current supplied by the battery is I = V/ R = (120 V)/(60 O) = 2 A, so
it must supply energy at a rate of P = IV = (2 A)(120 V) = 240 W.
(b) Since three times as much current will flow through the 40 O resistor as through the
branch containing 120 O of resistance, the current through the 20 O and 100 O resistors
must be 0.5 A.
(c) (i) V
a
 - V
b
 = IR
20
 + IR
100
 = (0.5 A)(20 O) + (0.5 A)(100 O) = 60 V.
(ii) Point a is at the higher potential (current flows from high to low potential).
(d) Because energy is equal to power multiplied by time, we get
E = Pt = I
2
Rt = (0.5 A)
2
(100 O)(10 s) = 250 J
(e) Using the equation R = ?L/ A, with A = p r
2
, we find
2. (a) The initial current, I
0
, is e/ r.
(b) Apply the equation Q(t) = Q
f
(1 - e
- t/ rC
). We want 1 - e
- t/ rC
 to equal 1/2, so
(c) Because it is connected to the positive terminal of the battery, the top plate will become
positively charged.
(d) When the current through r is zero, the capacitor is fully charged, with the voltage across
its plates matching the emf of the battery. Therefore,
(e) The current established by the discharging capacitor decreases exponentially according to
the equation I( t) = I
0
e
- t/t
 = (e/ R)e
- t/ RC
.
(f) The power dissipated is given by the joule heating law, P = I
2
R:
(g) We give two solutions. First, the total energy dissipated by the resistor will equal the
integral of P(t) from t = 0 to t = 8:
Page 4


Section II: Free Response
1. Consider the following circuit:
(a) At what rate does the battery deliver energy to the circuit?
(b) Find the current through the 20 O resistor.
(c) (i) Determine the potential difference between points a and b.
(ii) At which of these two points is the potential higher?
(d) Find the energy dissipated by the 100 O resistor in 10 s.
(e) Given that the 100 O resistor is a solid cylinder that’s 4 cm long, composed of a material
whose resistivity is 0.45 O m, determine its radius.
2. The diagram below shows an uncharged capacitor, two resistors, and a battery whose emf is e.
The switch S is turned to point a at time t = 0.
(Express all answers in terms of C, r, R, e, and fundamental constants.)
(a) Determine the current through r at time t = 0.
(b) Compute the time required for the charge on the capacitor to reach one-half its final value.
(c) When the capacitor is fully charged, which plate is positively charged?
(d) Determine the electrical potential energy stored in the capacitor when the current
through r is zero.
When the current through r is zero, the switch S is then moved to Point b; for the following
parts, consider this event time t = 0.
(e) Determine the current through R as a function of time.
(f) Find the power dissipated in R as a function of time.
(g) Determine the total amount of energy dissipated as heat by R.
Section II: Free Response
1. (a) The two parallel branches, the one containing the 40 O resistor and the other a total of
120 O, is equivalent to a single 30 O resistance. This 30 O resistance is in series with the
three 10 O resistors, giving an overall equivalent circuit resistance of 10 + 10 + 30 + 10 =
60 O. Therefore, the current supplied by the battery is I = V/ R = (120 V)/(60 O) = 2 A, so
it must supply energy at a rate of P = IV = (2 A)(120 V) = 240 W.
(b) Since three times as much current will flow through the 40 O resistor as through the
branch containing 120 O of resistance, the current through the 20 O and 100 O resistors
must be 0.5 A.
(c) (i) V
a
 - V
b
 = IR
20
 + IR
100
 = (0.5 A)(20 O) + (0.5 A)(100 O) = 60 V.
(ii) Point a is at the higher potential (current flows from high to low potential).
(d) Because energy is equal to power multiplied by time, we get
E = Pt = I
2
Rt = (0.5 A)
2
(100 O)(10 s) = 250 J
(e) Using the equation R = ?L/ A, with A = p r
2
, we find
2. (a) The initial current, I
0
, is e/ r.
(b) Apply the equation Q(t) = Q
f
(1 - e
- t/ rC
). We want 1 - e
- t/ rC
 to equal 1/2, so
(c) Because it is connected to the positive terminal of the battery, the top plate will become
positively charged.
(d) When the current through r is zero, the capacitor is fully charged, with the voltage across
its plates matching the emf of the battery. Therefore,
(e) The current established by the discharging capacitor decreases exponentially according to
the equation I( t) = I
0
e
- t/t
 = (e/ R)e
- t/ RC
.
(f) The power dissipated is given by the joule heating law, P = I
2
R:
(g) We give two solutions. First, the total energy dissipated by the resistor will equal the
integral of P(t) from t = 0 to t = 8:
Alternatively, simply notice that all the energy stored in the capacitor will be dissipated as
heat by the resistor R. But from part (d), we know that the initial energy stored in the
capacitor (before discharging) was Ce
2
.