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  • 天同
    Lv 7
    8 years ago
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    1. Since the voltage rating of each bulb is 3 volt, and is connected to 1 6 volt supply, we need to have two bulbs connected in series (to form a branch) to enable each bulb to be fully lit.

    Power supplied by the 6-volt source = 6 x 2 w = 12 w

    Each bulb consumes 1.5 w, hence each branch (consists of 2 bulbs in series) consumes 1.5 x 2 w = 3 w.

    No. of branches that can be connected to the power source in parallel

    = 12/3 = 4

    But there are 2 bulbs in each branch, the total no. of bulbs

    = 4 x 2 = 8

    2. (a) Let c be the critical angle

    sin(c) = 1/1.36

    hence, c = 47.33 degrees

    (b) Let the refracted angle at the other side (inside the soap) at A be r

    r = (c - 10) degrees = (47.33 - 10) degrees = 37.33 degrees

    Hence, sin(theta)/sin(37.33) = 1.36

    sin(theta) = 0.8247

    angle (theta) = 55.56 degrees

    (c)(i) Using Law of Reflection at the surface at B

    Reflected angle = c = 47.33 degrees

    Hence, incident angle of the reflected ray at the boundary at C

    = (47.33 + 10) degrees = 57.33 degrees

    2013-01-30 22:42:48 補充:

    Your question in supplementary:

    Current flows from high to low potential in an EXTERNAL CIRCUIT,but from low to high potential inside an electrical power source (like a battery or a generator/dynamo). Here,the moving coil acts like a generator producing electric current by electromagnetic induction.

    2013-01-30 22:46:43 補充:

    ... In the coil, the line AD acts as an external circuit, where as the line BC acts like an electrical generator (power source). Hence, D is higher in potential than A, but C is higher in potnetial than B.

    2013-01-30 22:49:16 補充:

    ... Current then flows from high to low potential in external circuit (from D to A) and from low to high potential in a power source (from B to C), thus completing a circuit.

    2013-01-30 22:52:42 補充:

    Just think of a case in which a resistor is connected to a battery. Current flows from the +ve to -ve pole of the battery (from high to low potential) through the resistor (external circuit), but from the -ve to +ve pole inside the battery (low to high potential).

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