1. A copper wire 3.2 mm in diameter, carries a 5.0-A current. What is the drift velocity of the free electrons?
    Solution.
    There is one free electron per atom. The atomic mass of Cu is 63.5 u, so 63.5 g of Cu contains one mole or 6.02 x 1023 free electrons. The mass density of copper is ρ =8.9 x 103 kg/m3, where ρ =m/V. So the number of free electrons per unit volume is
    n = N/V = N/(m/ρ) = N(I mol)/m(1 mol) ρ
        = [(6.02 x 1023 electrons/mol)/(63.5 x 10-3 kg)](8.9 x 103 kg/m3)
        = 8.4 x 1028 electrons/m3.
    Find the drift velocity of the free electrons.
    I = nevDA;
    5.0 A = (8.4 x 1028 electrons/m3)(1.602 x 10-19 C)vD(¼π)(3.2 x 10-3 m)2,
    vD = 4.6 x 10-5 m/s.
  2. A length of aluminum wire is connected to a precision 10.00-V power supply, and a current of 0.4212 A is precisely measured at 20.0ºC. The wire is placed in a new environment of unknown temperature where the measured current is 0.3618 A. What is the unknown temperature?.
    Solution.
    For the wire we have R = V/I. Find the temperature from
    R = R0(1 + αΔT);
    (V/I) = (V/I0)(1 + αΔT);
    (10.00 V/0.3618 A) = (10.00 V/0.4212 A){1 + [0.00429 (Cº)-1](T - 20ºC)},
    T = 58.3ºC.
  3. For the circuit shown in Figure, find the current through and the potential difference across each resistor.

    Solution.Find the equivalent resistance between a and b.
    R|| = (1/16 Ω + 1/48 Ω + 1/4 Ω)-1  = 3 Ω
    Rab = 3 Ω + 3 Ω = 6 Ω
    I = (12 V)/(6 Ω) = 2 A
    Vac = (3 Ω)(2 A) = 6 V,
    Vcb = (3 Ω)(2 A) = 6 V
    I1 = (6 V)/(16 Ω) =  0.375 A
    I2 = (6 V)/(48 Ω) =  0.125 A
    I3 = (6 V)/(4 Ω) =  1.5 A
  4. The 10 Ω resistor in Figure dissipating 40 W of power. How much power are the other two resistors dissipating?

    Solution.
    First find the current passing through 10-Ω resistor:
    I10Ω = (P/R)1/2 = [(40W)/(10 Ω)]1/2 = 2 A.
    Then find the voltage across 10- Ω resistor:
    V = I10Ω R = (2 A)(10 Ω) = 20 V.
    This voltage is also the voltage across the 20-Ω resistor.
    Find the power dissipated in 20-Ω resistor:
    P = V2/R = (20 V)2/(20 Ω) = 20 W.
    Find the current passing through 20-Ω resistor:
    I20Ω = V/R = (20 V)/(20 Ω) = 1 A.
    Find the current passing  5- Ω resistor:
    I = I10Ω + I20Ω = 2 A + 1 A = 3 A.
    Find the power dissipated in 5- Ω resistor:
    P = I5Ω 2 R = (3 A)2(5 Ω) = 45 W.