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3 years ago

What current is taken by the lamp when it is connected to a 220 v mains supply? what is the resistance of the lamp

Physics

1 answer:

eimsori [14]3 years ago

7 0

Answer:

Given parameters

Potential difference (V) = 220 V

Power of lamp 1 (P1) = 100 W

Power of lamp 2 (P2) = 60 W

Both the bulbs are connected in parallel. Therefore, the potential difference across each of them will be 220 V, because no division of voltage occurs in a parallel circuit.

We know that

Power = Voltage x Current

P = V x I

I = P/V

Lamp 1:

I1 = P1/V

I1 = 100/220

I1 = 5/11 A

Lamp 2:

I2 = P2/V

I2 = 60/220

I2 = 3/11 A

Net current (I) = I1 + I2

I = 5/11 + 3/11

I = 8/11

I = 0.7273 A

Hence, current drawn from the line is 0.727 A

Explanation:

Hope this helps!

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A motorbike accelerates from 15m/s to 25m/s in 15 seconds.

RSB [31]

distance traveled by a uniformly accelerated bike is given as

$d = \frac{v_f + v_i}{2} (t)$

here we know that

$v_f = 25 m/s$

$v_i = 15 m/s$

$t = 15 s$

now we will have from above equation

$d = \frac{15 + 25}{2} (15)$

$d = 20 (15) = 300 m$

so it will cover the total distance of 300 m

5 0

3 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

3 years ago

Please help ASAP will mark brainliest How do we find the amount of uncertainty we should expect in our measurement of the period

kicyunya [14]

Answer:

The relative uncertainty gives the uncertainty as a percentage of the original value. Work this out with: Relative uncertainty = (absolute uncertainty ÷ best estimate) × 100%. So in the example above: Relative uncertainty = (0.2 cm ÷ 3.4 cm) × 100% = 5.9%. The value can therefore be quoted as 3.4 cm ± 5.9%.

Explanation:

hope it helps :)

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3 years ago

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