What happens to the resistance of two resistors when connected in series circuit

2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from

OLEGan [10]

Answer:

$Relative\ Velocity = 105m/s$

Explanation:

Given

$V_A = 45m/s$

$V_B = 60m/s$

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a $sum\ of\ velocities\ of$ both trains:

This is shown below:

$Relative\ Velocity = V_A + V_B$

$Relative\ Velocity = 45m/s + 60m/s$

$Relative\ Velocity = 105m/s$

4 0

2 years ago

A fixed 15.3-cm-diameter wire coil is perpendicular to a magnetic field 0.77 T pointing up. In 0.20 s , the field is changed to

maksim [4K]

Answer:

Induced emf will be 0.468 volt

Explanation:

We have given diameter of wire d = 15.3 cm

So radius $r=\frac{d}{2}=\frac{15.3}{2}=7.65cm$

So area $A=\pi r^2=3.14\times (7.65)^2=183.76\times 10^{-4}m^2$

Change in magnetic field dB = 0.26 - 0.77 = -0.51 T

Time for change in magnetic field dt = 0.26 sec

We know that emf is given by $e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-183.76\times 10^{-4}\times \frac{-0.51}{0.2}=0.0468volt$

5 0

2 years ago

When rockets split do they come back to earth?

attashe74 [19]

I'm not 100% sure, but I believe what you mean is when they eject the old propulsion motors. Yes, they land in the ocean and the US Navy retrieves them for later use.

4 0

2 years ago

Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show

ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = $\frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}$

We know that the inductance of the coil is given by the formula:

$L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH$

4 0

2 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

8 0

2 years ago

What happens to the resistance of two resistors when connected in series​ circuit

What happens to the resistance of two resistors when connected in series circuit