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Q30P

Expert-verifiedFound in: Page 937

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A ${\mathbf{50}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\Omega}}$ resistor is connected, as in Figure to an ac generator with ${{\mathbf{\in}}}_{{\mathbf{m}}}{\mathbf{=}}{\mathbf{30}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{V}}$****.**

**(a) What is the amplitude of the resulting alternating current if the frequency of the emf is ${\mathbf{1}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{kHz}}$? (b) What is the amplitude of the resulting alternating current if the frequency of the emf is ${\mathbf{8}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{kHz}}$?**

**A resistor is connected across an alternating-current generator.**

- The amplitude of the resulting alternating current if the frequency of the emf has value $f=1.00\mathrm{kHz}$ is $0.600\mathrm{A}$.
- The amplitude of the resulting alternating current if the frequency of the emf has value $f=8.00\mathrm{kHz}$ is $0.600\mathrm{A}$.

- Emf of the ac generator, ${\in}_{m}=30.0\mathrm{V}$
- Series resistance $R=50\mathrm{\Omega}$

**Using Ohm’s law and substituting the given values of the amplitude of the ac generator and resistance of the resistor, we can find the amplitude of the alternating current. As the resistance of the circuit is independent of frequency, the current value remains the same for the same resistor.**

The voltage equation from Ohm’s law,

${i}_{c}=\frac{{\in}_{m}}{R}$ (i)

Here, $R$ is the resistance of the circuit and ${\in}_{m}$ the emf applied across the circuit.

The amplitude of alternating current if the frequency of emf is $1\mathrm{kHz}$.

Using the given data in equation (i), the current through the resistor is given as:

$I=\frac{30\mathrm{V}}{50\mathrm{\Omega}}\phantom{\rule{0ex}{0ex}}I=0.600\mathrm{A}$

Hence, the value of the current is $0.600\mathrm{A}$.

The amplitude of alternating current if the frequency of emf is $8\mathrm{kHz}$.

The current is independent of frequency. So, it will be the same as calculated in part (a) above.

Hence, the value of the current is $0.600\mathrm{A}$.

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