Physics, asked by ssnn2, 7 months ago

A 120 V, 20 Hz source is connected to a series circuit consisting of 5 ohm capacitive reactance, a 1.6 ohm resistor
and a coil with resistance and inductive reactance of 3-ohm and 1.2-ohm respectively. Calculate (a) the input
impedance (b) the circuit current

Answers

Answered by TrishaNikhilJaiswal
4

Answer:

When alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other’s effect. Figure 1 shows an RLC series circuit with an AC voltage source, the behavior of which is the subject of this section. The crux of the analysis of an RLC circuit is the frequency dependence of XL and XC, and the effect they have on the phase of voltage versus current (established in the preceding section). These give rise to the frequency dependence of the circuit, with important “resonance” features that are the basis of many applications, such as radio tuners.

The figure describes an R LC series circuit. It shows a resistor R connected in series with an inductor L, connected to a capacitor C in series to an A C source V. The voltage of the A C source is given by V equals V zero sine two pi f t. The voltage across R is V R, across L is V L and across C is V C.

Figure 1. An RLC series circuit with an AC voltage source.

The combined effect of resistance R, inductive reactance XL, and capacitive reactance XC is defined to be impedance, an AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an RLC circuit are related by an AC version of Ohm’s law:

I

0

=

V

0

Z

or

I

rms

=

V

rms

Z

.

Here I0 is the peak current, V0 the peak source voltage, and Z is the impedance of the circuit. The units of impedance are ohms, and its effect on the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression for Z in terms of R , XL, and XC, we will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled VR, VL, and VC in Figure 1. Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in R, L, and C are equal and in phase. But we know from the preceding section that the voltage across the inductor VL leads the current by one-fourth of a cycle, the voltage across the capacitor VC follows the current by one-fourth of a cycle, and the voltage across the resistor VR is exactly in phase with the current. Figure 2 shows these relationships in one graph, as well as showing the total voltage around the circuit V = VR + VL + VC, where all four voltages are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage around the circuit V is also the voltage of the source. You can see from Figure 2 that while VR is in phase with the current, VL leads by 90º, and VC follows by 90º. Thus VL and VC are 180º out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are not aligned (not in phase), the peak voltage V0 of the source does not equal the sum of the peak voltages across R, L, and C. The actual relationship is

V

0

=

V

0

R

2

+

(

V

0

L

V

0

C

)

2

,

where V0R, V0L, and V0C are the peak voltages across R, L, and C, respectively. Now, using Ohm’s law and definitions from Reactance, Inductive and Capacitive, we substitute V0 = I0Z into the above, as well as V0R = I0R, V0L = I0XL, and V0C = I0XC, yielding

I

0

Z

=

I

0

2

R

2

+

(

I

0

X

L

I

0

X

C

)

2

=

I

0

R

2

+

(

X

L

X

C

)

2

I0 cancels to yield an expression for Z:

Z

=

R

2

+

(

X

L

X

C

)

2

,

Explanation:

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Answered by amitkumar9266
0

Answer:

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