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# Current Electricity:Variable voltage, current in L, C and R

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Determine and plot as a function of time the current through a component if the voltage across it has the waveform shown in Figure and the component is a:

a) Resistor R = 7 &#937;
b) Capacitor C = 0.5 &#956;F
c) Inductor L = 7 mH

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Determine and plot as a function of time the current through a component if the voltage across it has the waveform shown in Figure P4.17 and the component is a:
a) Resistor R = 7 Ω
b) Capacitor C = 0.5 μF
c) Inductor L = 7 mH

a)

In case of a resistor the Ohm's law as gives the current

I = V/R

For the first interval of time 0-5 ms the current is increasing with voltage from
I = 0 at t = 0
to I = V/R = 15/7 = 2.143 A
linearly and then remains constant with voltage for next 5 ms. After that as the voltage becomes zero the current becomes zero and hence the graph is given as shown.

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b)

(Here no resistance is given means we have to consider a pure capacitance)

The potential difference across a capacitor is given by V = q/C where q is charge on the capacitor. As the voltage increases, charge q on the capacitor will increase and the hence there through the circuit is given as

I = dq/dt = d (CV)/dt = C dV/dt

For the first interval of 5 ms the voltage is changing at a constant rate

dV/dt = 15/(5*10-3) = 3000 V/s

and hence the current is constant given by

I = C (dV/dt)
= 0.5*10-6*3000 = 1500*10-6 A
= 1500 A

For the second interval of time the voltage is constant and hence the rate of change voltage dV/dt is zero and no current will flow through the circuit.

At t = 10 ms the voltage across the capacitor becomes zero suddenly, means the charge on the capacitor suddenly becomes zero. The charge already stored in the capacitor will flow back in an infinitely short time (as no resistance is there) and after that the current becomes zero. That pulse of negative current cannot be shown on the graph as the time interval is infinitely small as compare to the given scale. The graph is shown as in figure.

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c)

The voltage across the inductor is equal to the induced emf because of change in current in it.

Faraday's law gives the relation between the induced emf and the current in an inductor as

In the circuit for the loop net voltage must be zero hence V - e = 0 (Krichoffs's law}

This gives

Or

Or

Now the voltage is changing with time linearly from t = 0 to 5 ms, and thereafter it is constant. Hence the during the first interval
Volts/s.

Hence the voltage across the inductor as a function of time t, in the first interval of time is given by

Hence the current at the end of first interval is given by using equation (1) as

The current is square function of time the curve will be a parabola.

Now the voltage in the second interval is constant (15 V) hence the current I2 increases linearly and at time t = 10 s is given by using the same equation (1) as

A
Hence the current at 10 ms is given by

I2 = I1 + 10.714
= (5.357 + 10.714)
= 16.071 A.

After 10 ms the voltage across the inductor is suddenly becomes zero hence induced emf in the inductor is zero, the rate of change of the current is zero and the current remains constant thereafter = 16.071 A.

The graph is shown in the figure.

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This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com October 4, 2022, 12:33 am ad1c9bdddf>
https://brainmass.com/physics/emf/current-electricity-variable-voltage-current-70302