Physics

A circuit element X when connected to an a.c. supply of peak voltage 100V given a peak current of 5A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

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Contributor-Level 10

v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$ &nb

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v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$

$5 = \frac{100}{R_{y}} \Rightarrow R_{y} = \frac{100}{5} = 20 Ω$

When x and y both are connected in series :-

v = 100sin $ω t$

tanθ = 1 = q = 45°

R₀ = $\sqrt[]{R_{x}^{2} + R_{y}^{2}} = 20 \sqrt[]{2} Ω$

$∴ \overset{o}{l_{0}} = \frac{v_{0}}{R_{0}} = \frac{100}{20 \sqrt[]{2}} = \frac{5}{\sqrt[]{2}} A .$

$∴ l_{m s} = \frac{l_{0}}{\sqrt[]{2}}$

$= \frac{5}{\sqrt[]{2} \cdot \sqrt[]{2}}$

$= \frac{5}{2} A = \frac{5}{2} A$

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5 months ago

The equation of current in a purely inductive circuit is 5sin(49pt – 30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be: ${L e t π = \frac{22}{7}}$

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Physics Physics Moving Charges and Magnetism

Contributor-Level 10

$i = 5 s i n (49 π t - 30 °)$

$L = 30 m H, π = \frac{22}{7}$

$X_{L} = ω L = 49 π * 30 * 1 0^{- 3}$

$= 49 \frac{22}{7} * 30 * 1 0^{- 3}$

$V_{L} = 23.1 s i n (49 π t + 60 °)$

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5 months ago

As shown in the figure, a metallic rod of linear density 0.45 km^-¹ is lying horizontally on a smooth incline plane which makes an angle of 45° 5with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be:

{Use g = m/s²}

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Physics Physics Moving Charges and Magnetism

Contributor-Level 10

Linear density, $λ =$ 0.45 kg/m

Let length = $l$ $∴$ = m = 0.45 $l$

B = 0.15 T

For equilibrium of rod :-

mg sin 45° = FB sin 45°

$\Rightarrow (0.45 l) g = l l B$

So, $l = \frac{0.45 * 10}{0.15} = 30 A$ $\frac{}{}$

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5 months ago

The current sensitivity of a galvanometer can be increased by:

(a) decreasing the number of turns

(b) increasing the magnetic field

(c) decreasing the area of the coil

(d) decreasing the torsional constant of the spring

Choose the most appropriate answer from the options given below:

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Physics Physics Motion in Straight Line

Contributor-Level 10

Current sensitivity, $\frac{}{}$ $S i = \frac{N A B}{k}$

$N i A B = k θ$

$θ = (\frac{N A B}{k}) i$

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5 months ago

A juggler throws ball vertically upwards with same initial velocity in air. When the first ball reaches its highest positions he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

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Contributor-Level 10

Time to reach at max height t = $\frac{u}{g} n o .$ of balls thrown in 1 sec = n.

So, time taken by each ball to reach max^m height, = $\frac{1}{n} s e c$

i.e. $\frac{u}{g} = \frac{1}{n} \Rightarrow u = \frac{g}{n}$ So, h_max = $\frac{u^{2}}{2 g}$

= $\frac{g^{2}}{2 g n^{2}}$

$= \frac{g}{2 n^{2}}$

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5 months ago

A wire of resistance R₁ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

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Physics Physics Moving Charges and Magnetism

Contributor-Level 10

$R_{2} = ρ \frac{2 l}{A / 2} = \frac{4 ρ l}{A} = 4 R_{1}$
$\frac{R_{2}}{R_{1}} = \frac{4}{1} \to 4 : 1$

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5 months ago

A wire X of length 50cm carrying a current of 2A is placed parallel to a long wire Y of length 5m. The wire Y current of 3A. The distance between two wires is 5cm and currents flow in the same direction. The force acting on the wire Y is

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Physics Electrostatic Potential and Capacitance

Contributor-Level 10

F_x_-y = $\frac{μ_{0} i_{1} i_{2}}{2 π r} * (. 5)$

$= \frac{2 * 1 0^{- 7} * 6 * 0.5}{0.05}$

$= \frac{6}{5} * 1 0^{- 5} N = 1.2 * 1 0^{- 5}$

Force on Y = 1.2 * 10^-5N towards 'x'

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5 months ago

Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of constant K such that the equivalence capacitance of the system became $24 μ F$ . The value of K will be:

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Contributor-Level 10

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{40 * 40 k}{40 (1 + k)} = 24$

$\begin{array}{l} \Rightarrow 40 k = 24 + 24 k \\ \Rightarrow 16 k = 24 \end{array}$

$k = \frac{24}{16} = \frac{3}{2} = 1.5$

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5 months ago

A 1m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.

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Contributor-Level 10

R_w = 2R_y

$\Rightarrow ρ (\frac{2 x}{\frac{A}{2}}) = \frac{2 ρ (1 - x)}{A}$

$\frac{4 ρ x}{A} = \frac{2 ρ (1 - x)}{A}$

->4x = 2 – 2x Þ 6x = 2

x = $\frac{2}{6} = \frac{1}{3}$ $∴ \frac{L_{x}}{L_{y}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{3} * \frac{3}{2}$

$= \frac{1}{2}$

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5 months ago

Physics Waves

In the wave equation

$y = 0.5 s i n \frac{2 π}{λ} (400 t - x) m$

the velocity of the wave will be:

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