Answer to Question #101119 in Electric Circuits for Fareedah

Question

1. A rectangular coil measuring 200 mm by 100 mm is mounted such
that it can be rotated about the midpoints of the 100 mm sides.
The axis of rotation is at right angles to a magnetic field of uniform
flux density 0.05 T. Calculate the flux in the coil for the following
conditions:
(a) the maximum flux through the coil and the position at which
it occurs;
(b) the flux through the coil when the 100 mm sides are inclined
at 45° to the direction of the flux

2. Three resistors of 6 Ω, 9 Ω and 15 Ω are connected in
parallel to a 9 V supply. Calculate: (a) the current in
each branch of the network; (b) the supply current;
(c) the total effective resistance of the network.

Answer 1

1. Magnetic flux through a flat coil $\Phi = B A \cos\alpha$ ,

where:

$\quad B \text{ - flux density, T} \\ \quad A = w * h \text{ - area of the coil, }m^2 \\ \quad \alpha \text{ - angle between the magnetic induction vector } \\ \text{ and the normal to the plane of the coil } \\ \quad w, h \text{ - width and height of the coil, m }$

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

$\quad I_{6\Omega} = 9 / 6 = 1.5 \text{ (A)} \\ \quad I_{9\Omega} = 9 / 9 = 1 \text{ (A)} \\ \quad I_{15\Omega} = 9 / 15 = 0.6 \text{ (A)}$

2.b Supply current

$\quad I_\Sigma = I_{6\Omega} + I_{9\Omega} + I_{15\Omega} = 1.5 + 1 + 0.6 = 3.1 \text{ (A)}$

2.c Total effective resistance of the network

$\quad R_\Sigma = \cfrac V I_\Sigma = 9 / 3.1 \approx 2.9 \text{ (Ohm)}$