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amm1812
3 years ago
13

What are the advantages and disadvantages of series and parallel circuits?

Physics
1 answer:
DerKrebs [107]3 years ago
6 0
Parallel circuit

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 circuit

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 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
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Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
2 years ago
Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th
Delvig [45]

Explanation:

As the given data is as follows.

    R_{1} = 680 \ohm ohm\ohm,    R_{2} = 720 \ohm ohm,

   R_{3} = 1.2 k\ohm = 1200 \ohm   (as 1 k ohm = 1000 m)

(a)   We will calculate the maximum resistance by combining the given resistances as follows.

      Max. Resistance = R_{1} + R_{2} + R_{3}

                                  = (680 + 720 + 1200) \ohm ohm

                                  = 2600 ohm

or,                               = 2.6 k\ohm ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 k\ohm ohm.

(b)   Now, the minimum resistance is calculated as follows.

      Min. Resistance = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

                                 = \frac{1}{680} + \frac{1}{720} + \frac{1}{1200}

                                 = 3.683 \times 10^{-3} ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is 3.683 \times 10^{-3} ohm.

3 0
3 years ago
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