Answer in Electric Circuits for Fareedah #101119
1. Magnetic flux through a flat coil “Phi = B A cosalpha” ,
where:
“quad B text{ – flux density, T} \nnquad A = w * h text{ – area of the coil, }m^2 \nnquad alpha text{ – angle between the magnetic induction vector } \ntext{ and the normal to the plane of the coil } \nquad w, h text{ – width and height of the coil, m }”
1.a The maximum flux through the coil occurs when the plane of the coil is perpendicular to the magnetic field.
“quad Phi_{max} = 0.05 * 0.2 * 0.1 * cos 0 = 10^{-3} text{ (Wb)}”
1.b
“quad Phi_{45^circ} = 0.05 * 0.2 * 0.1 * cos {45^circ} approx 7.07 * 10^{-4} text{ (Wb)}”
2. Will use Ohm’s law “I = cfrac V R”
2.a Current in each branch of the network
“quad I_{6Omega} = 9 / 6 = 1.5 text{ (A)} \nquad I_{9Omega} = 9 / 9 = 1 text{ (A)} \nquad I_{15Omega} = 9 / 15 = 0.6 text{ (A)}”
2.b Supply current
“quad I_Sigma = I_{6Omega} + I_{9Omega} + I_{15Omega} = 1.5 + 1 + 0.6 = 3.1 text{ (A)}”
2.c Total effective resistance of the network
“quad R_Sigma = cfrac V I_Sigma = 9 / 3.1 approx 2.9 text{ (Ohm)}”
