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3 years ago

How to increase the current in a circuit

Physics

1 answer:

ivann1987 [24]3 years ago

3 0

Answer:

So to increase current of the circuit what you can do is :

1. Use conductor of low resistivity, ¶.

2. Use conductor of small length.

3. Use thick wire.

4. Decrease the temperature of the circuit.

5. If operating temprature is high than use semiconductor, because it have negative temprature coefficient.

6. Minimise the circuit losses.

You might be interested in

How many times larger than a centigram is a dekagram

Umnica [9.8K]

Answer:

A dekagram is thousand (1000) times larger than a centigram.

Explanation:

→ [1 dekagram = 1,000 centigrams]

→ 1 dekagram = 10 grams

→ 10 grams = 100 decigrams

→ 100 decigrams = 1,000 centigrams

3 0

3 years ago

Many of the whales in the ocean rely upon tiny marine organisms, called plankton, for food. If all of the plankton suddenly died

exis [7]

Answer:

I believe the answer to be B.

Explanation:

Without food, the whales would die.

8 0

3 years ago

A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays

IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

$p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s$

Final momentum

$p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s$

4 0

3 years ago

When current first leaves the battery, how much voltage does it have

rjkz [21]

Answer:

depends on the voltage of battery

Explanation:

for example if you connect a battery of 6V,6V will be provided

8 0

3 years ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

3 0

3 years ago