Chapter 20.
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2). Current is the amount of charge per unit time.
I = q/t
The total charge that pass through the CDROM in 1 minute is
q = I * t = 0.27 A * 60 s = 16.2 C
The charge of one electron is 1.6*10^(-16) C, thus the number of
electrons is given by a ratio of the total chagre to charge of one
electron:
N = q/1.6*10^(-19) C = 16.2 C/1.6*10^(-19) C = 10^(20)
3). Use the Ohm's law:
I = V / R = 120 V / 14 Ohm = 8.57 A
6). The electric power is P = I * V
The energy delivered to the battery is equal to work performed
by using power for a certain time
E = P * t = 6 A * 12 V * 5 hours =
= 6 A * 12 V * 18000s = 1.296 * 10^6 J
35). Delivered electric energy is
E = P * t
We must express power in terms of quantities available in the problem
P = R * I^2
Thus
E = R * I^2 * t = 5.3 Ohm * (25 A)^2 * 31 days * 9 hours/day =
= 9.2 * 10^5 W*hour = 9.2*10^2 kW*hour
The total cost of running the furnace is
Cost = 920 kWh * $0.1 /kWh = $92
42). First we need to calculate the current in the circuit
I = V / R_tot
The total resistance of the two resistors connected in
series is equal to a sum of individual resistances
R_tot = 36 Ohm + 18 Ohm = 54 Ohm
I = 15 V / 54 Ohm = 0.28 A
The voltage across the 18 Ohm resistor is
V_18 = I * R = 18 Ohm * 0.28 A = 5 V
The voltage across 36 Ohm resistor is
V_36 = I * R = 36 Ohm * 0.28 A = 10 V
51). For N resistors each with resistance R and all connected
in parallel the total resistance of a system is
1/R_tot = 1/R + 1/R + 1/R +....1/R = N/R
---> N = R / R_tot = 4 Ohm / (1/16 Ohm) = 64
We need 64 resistors.
77). Apply the Kirchhof's rules.
The current in the circuit is given by the equation constructed by adding
all potential drops following the loop. The starting point is arbitrary.
I start at point A and follow the loop clockwise
30 V - 27 Ohm * I - 10 V - 12 Ohm * I - 8 Ohm * I - 5 Ohm * I = 0
20 V - 52 Ohm * I = 0
a) I = 20 V / 52 Ohm = 0.38 A
The answer is positive and that means that the current flows in the
direction in which we formed the equation i.e., clockwise.
b) Potential difference between points A and B is given by the battery
potential decreased by the potential drop across the 27 Ohm resistor
V_B - V_A = 30V - 0.38 A * 27 Ohm = 19.7 V
c) The potential difference is positive
V_B - V_A > 0 ====> V_B > V_A
The potential at point B is higher than potential at point A.
90). Total charge on a capacitors (see formula 19.8) is
q = V * C_tot
The total capacitance of two capacities connected in parallel is
C_tot = C_1 + C_2 = 2*10^(-6) F + 2*10^(-6) F = 6*10^(-6) F
For parallel connction
q = 60 V * 6*10^(-6) F = 3.6*10^(-4) C
The total capacitance of two capacities connected in series is
1/C_tot = 1/C_1 + 1/C_2
C_tot = 1/(1/C_1 + 1/C_2) = 1/(1/4 + 1/6) microF =
= 2.4 microF =>
For connection in series
q = 60 V * 2.4*10^(-6) F = 1.44*10^(-4) C