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Question: Reading of the voltmeter is \({V_1} = 40V\), \({V_2} = 40V\) and \({V_3} = 10V\). Find: ![](https...

Reading of the voltmeter is V1=40V{V_1} = 40V, V2=40V{V_2} = 40V and V3=10V{V_3} = 10V. Find:

1. i0,irms,i(t){i_0}, {i_{rms}}, i(t)
2. ε0{\varepsilon _0}
3. LL and CC

Explanation

Solution

All three are in series and the current in series is the same. This means the value of current should be the same through all the three elements. However the voltage across the capacitor lags from the voltage across the resistor and the voltage across the resistor lags from voltage across the inductor.

Complete step by step solution:
Although the speed of gaseous particles is continuously changing, the velocity distribution does not change. It is difficult to calculate the speed of each single particle, but often we reason for the average behaviour of the particles. Particles that travel in comparison have opposite sign speeds. Since a gas' particles are in random motion, it is plausible that there will be about as many moving in one direction as in the opposite direction, meaning that the average velocity for a collection of gas particles equals zero; as this value is unhelpful, the average of velocities can be determined using an alternative method.
irms=V1R{i_ {rms}} = \dfrac{{{V_1}}}{R}
Where, irms{i_ {rms}} is root mean square velocity
irms=404=10A\Rightarrow {i_ {rms}} = \dfrac{{40}}{4} = 10A
Formula for i0{i_0} which is the current produced by the voltage source.
i0=2irms{i_0} = \sqrt 2 {i_ {rms}}
i0=102A\Rightarrow {i_0} = 10\sqrt 2 A
ε0{\varepsilon _0} is the voltage.
ε0=E02V21+(V2V1)2{\varepsilon _0} = \dfrac{{{E_0}}}{{\sqrt 2}}\sqrt {{V^2} _1 + {{({V_2} - {V_1})} ^2}}
ε0=E02402+(4010)2\Rightarrow {\varepsilon _0} = \dfrac{{{E_0}}}{{\sqrt 2}}\sqrt {{{40} ^2} + {{(40 - 10)} ^2}}
ε0=50V\Rightarrow {\varepsilon _0} = 50V
Now,
XL=(ωL)=V2irms{X_L} = (\omega L) = \dfrac{{{V_2}}}{{{i_{rms}}}}
Where, XL{X_L} is Inductive Reactance
XL=4Ω\Rightarrow {X_L} = 4\Omega
L=4ω=4100πHL = \dfrac{4}{\omega} = \dfrac{4}{{100\pi}}H
L=125π\Rightarrow L = \dfrac{1}{{25\pi}}
Where, XC{X_C} is Capacitive Reactance
XC=ωC=V1irms{X_C} = \omega C = \dfrac{{{V_1}}}{{{i_{rms}}}}
XC=1Ω\Rightarrow {X_C} = 1\Omega
C=1ω=1100πF\Rightarrow C = \dfrac{1}{\omega} = \dfrac{1}{{100\pi}}F

Note: By squaring speeds and taking the square root, we conquer the 'directional' component of speed and at the same time achieve the average speed of the particles. As the value lacks the trajectory of the particles, the average speed is now referred to as the value. Root-medium-square velocity is the calculation of particle speed in a gas described by the square root of the average molecular velocity square of a gas.