Question

24. An AC source operating at 60. Hz with a maximum voltage of 170 V is connected in series with a resistor (R 1.2 k2) and an inductor (L = 2.8 H). (a) what is the maximum value of the current in the circuit? (b) What are the maximum values of the potential difference across the resistor and the induc- tor? (c) When the current is at a maximum, what are the mag- nitudes of the potential differences across the resistor, the inductor, and the AC source? (d) When the current is zero, what are the magnitudes of the potential difference across the resistor, the inductor, and the AC source? er A nerson is working near the secondary of a transformer, as

Answer 1

Hi,

Hope you are doing well.

PART (a):

We know that, the given circuit is a series LR circuit.

We have, Maximum value of current in the circuit is given by,

$I=\frac{V}{Z}$

Where,

is the maximum voltage across the circuit (V).

is the impedance of the circuit (ohms), which is given by,

$Z=\sqrt{R^{2}+(\omega.L)^{2}}$

Where,

is the resistance of the resistor (ohms).

$\omega$ is the frequency of the input source (Hz).

is the inductance of the inductor (H).

Given that,

Resistance of the resistor, $R=1.2\;k\Omega=1200\;\Omega$

Frequency of the input source, $\omega=60\;Hz$

Inductance of the inductor, $L=2.8\;H$

$\therefore$ Impedance of the circuit is given by,

$Z=\sqrt{R^{2}+(\omega.L)^{2}}=\sqrt{(1200\;\Omega)^{2}+(60\;Hz\times2.8\;H)^{2}}$

$\therefore Z=\sqrt{1468224}\;\Omega$

$\mathbf{\therefore Z=1211.7029\;\Omega}$

Also, Given that,

Maximum voltage across the circuit, $\mathbf{V=170\;V}$

$\therefore$ Maximum value of current in the circuit is given by,

$I=\frac{V}{Z}=\frac{170\;V}{1211.7029\;\Omega}$

$\mathbf{\therefore I=0.140298\;A}$

PART (b):

The maximum possible value of voltage across the resistor and inductor will be 170 V which will happen only in some special cases.

PART (c):

We know that, According to Ohm's law, voltage across a resistor is given by,

$V_{R}=I.R$

Where,

is the current through the resistor (A).

is the resistance of the resistor (ohms).

Given that,

Current through the resistor, $I=0.140298\;A$

Resistance of the resistor, $R=1.2\;k\Omega=1200\;\Omega$

$\therefore$ Voltage across a resistor is given by,

$V_{R}=I.R=0.140298\;A\times1200\;\Omega$

$\mathbf{\therefore V_{R}=168.3581\;V}$

Also, We know that, According to Ohm's law, voltage across a inductor is given by,

$V_{L}=I.X_{L}$

Where,

is the current through the inductor (A).

$X_{L}$ is the inductive reactance of the inductor (ohms) which is given by,

$X_{L}=L.\omega$

Where,

$\omega$ is the frequency of the input source (Hz).

is the inductance of the inductor (H).

Given that,

Frequency of the input source, $\omega=60\;Hz$

Inductance of the inductor, $L=2.8\;H$

$\therefore$ Inductive reactance of the inductor is given by,

$X_{L}=L.\omega=2.8\;h\times60\;Hz$

$\mathbf{\therefore X_{L}=168\;\Omega}$

Also, Given that,

Current through the inductor, $I=0.140298\;A$

$\therefore$ Voltage across a inductor is given by,

$V_{L}=I.X_{L}=0.140298\;A\times168\;\Omega$

$\mathbf{V_{L}=23.5701\;V}$

PART (d):

We know that, According to Ohm's law, voltage across a resistor is given by,

$V_{R}=I.R$

Where,

is the current through the resistor (A).

is the resistance of the resistor (ohms).

When the current through the resistor, $I=0\;A$ , then voltage across a resistor is given by,

$\mathbf{V_{R}=0\;A\times R=0\;V}$

Similarly,

Voltage across a inductor is given by,

$V_{L}=I.X_{L}$

Where,

is the current through the inductor (A).

$X_{L}$ is the inductive reactance of the inductor (ohms) which is given by,

$X_{L}=L.\omega$

When the current through the inductor, $I=0\;A$ , then voltage across the inductor is given by,

$\mathbf{V_{L}=0\;A\times X_{L}=0\;V}$

Hope this helped for your studies. Keep learning. Have a good day.

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