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Murrr4er [49]
3 years ago
13

By increasing the circuits resistance R, in a capacitive circuit the phase difference between the current in the circuit and pot

ential difference across the source.
A. Gets larger
B. Gets smaller
C. Remains unchanged
D. Depending on the values could get smaller or larger
E. None of these
Physics
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

B) Gets smaller

Explanation:

The difference of phase between current and voltage in a AC circuit is the phase  angle and it depends on the value of Z ( circuit impedance)

Z = R + X   where R is the resistive component and X the reactance component, which is due either to a presence of an  inductor or a capacitor. In any case the total impedance depends on R the resistive, and the phase angle φ is:  

tan⁻¹ φ =  X/R

Have a look to a pure capactive circuit (we are talking about AC current) in this case current leads voltage by 90⁰. If we add a resistor in the circuit  the current still will lead a voltage but in this condition the phase angle will be smaller,

If  R increase, X/R  decrease and tan⁻¹ φ also decrease

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Answer:

Explanation:

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The shades are very different
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How many moles of MgCl2 are there in 329 g of the compound?<br><br> Your Answer:
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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
4 years ago
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