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MatroZZZ [7]
2 years ago
13

1. Write the advantages and disadvantages of connecting lamps (a) in series and (b) in parallel. A3 2. a) 0.2A Draw Fig 5.30 int

o your exercise book and name the way in which the lamps are connected and answer questions b and c A₂ b) If A3 read 0.2A, what do the other two ammeters read? A1 c) If one of the lamps is removed explain what would happen to the other lamp. Fig 5.30 Reading of ammeter in series circuit​

Physics
1 answer:
spin [16.1K]2 years ago
7 0

Answer:

parallel connection:Advandages: 1. Every unit that is connected in a parallel circuit gets equal amount of voltage.2. It becomes easy to connect or disconnect a new element without affecting the working of other elements.3. If any fault happened to the circuit, then also the current is able to pass through the circuit through different paths.Disadvantages: 1. It requires the use of lot of wires.2. We cannot increase or multiply the voltage in a parallel circuit.3. Parallel connection fails at the time when it is required to pass exactly same amount of current through the units.series connection:Advantages: 1. Series circuits do not overheat easily. This makes them very useful in the case of something that might be around a potentially flammable source, like dry plants or cloth.2. Series circuits are easy to learn and to make. Their simple design is easy to understand, and this means that it’s simple to conduct repairs .3. we can add more power devices, they have a higher output in terms of voltage .4. The current that flows in a series circuit has to flow through every component in the circuit. Therefore, all of the components in a series connection carry the same current.Disadvantages: 1.If one point breaks in the series circuit,the total circuit will break.2. As the number of components in a circuit increases ,greater will be the circuit resistance.

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(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

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(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s)   ==>   <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

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A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu
ioda

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

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The magnetic flux is given as:

\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb

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