Take R = 1.50 Qk and C = 1.50 MF. A variable-frequency AC voltage source producing 18 Vrms is connected across the circuit, and an AC ammeter in series with the voltage source shows a minimum reading when the frequency of the applied voltage is 855 Hz. Find (a) the inductance and (b) that minimum value of the ammeter reading

Sure, let's solve the problem step by step.

Given:

• Resistance, $R = 1.50 \, \Omega$
• Capacitance, $C = 1.50 \, \mu F$
• RMS voltage, $V_{rms} = 18 \, V$
• Resonance frequency, $f = 855 \, Hz$

(a) To find the inductance $L$:

At resonance in an RLC circuit, the inductive reactance $X_L$ equals the capacitive reactance $X_C$.

$X_L = X_C$

The inductive reactance is given by:

$X_L = 2 \pi f L$

The capacitive reactance is given by:

$X_C = \frac{1}{2 \pi f C}$

Setting $X_L = X_C$:

$2 \pi f L = \frac{1}{2 \pi f C}$

Solving for $L$:

$L = \frac{1}{(2 \pi f)^2 C}$

Substitute the given values:

$L = \frac{1}{(2 \pi \times 855)^2 \times 1.50 \times 10^{-6}}$

$L = \frac{1}{(2 \pi \times 855)^2 \times 1.50 \times 10^{-6}}$

$L = \frac{1}{(5.366 \times 10^3)^2 \times 1.50 \times 10^{-6}}$

$L = \frac{1}{28.8 \times 10^6 \times 1.50 \times 10^{-6}}$

$L = \frac{1}{43.2}$

$L \approx 0.0231 \, H$

So, the inductance $L$ is approximately $0.0231 \, H$.

(b) To find the minimum value of the ammeter reading:

At resonance, the impedance $Z$ of the circuit is purely resistive and equals the resistance $R$.

$Z = R$

The current $I$ in the circuit can be found using Ohm's law:

$I = \frac{V_{rms}}{Z}$

Since $Z = R$:

$I = \frac{V_{rms}}{R}$

Substitute the given values:

$I = \frac{18 \, V}{1.50 \, \Omega}$

$I = 12 \, A$

So, the minimum value of the ammeter reading is $12 \, A$.

Sure, let's solve the problem step by step.

Given:

• Resistance, $R = 1.50 \, \Omega$
• Capacitance, $C = 1.50 \, \mu F$
• RMS voltage, $V_{rms} = 18 \, V$
• Resonance frequency, $f = 855 \, Hz$

(a) To find the inductance $L$:

At resonance in an RLC circuit, the inductive reactance $X_L$ equals the capacitive reactance $X_C$.

The resonance frequency $f$ is given by: $f = \frac{1}{2\pi\sqrt{LC}}$

Rearranging to solve for $L$: $L = \frac{1}{(2\pi f)^2 C}$

Substitute the given values: $L = \frac{1}{(2\pi \times 855)^2 \times 1.50 \times 10^{-6}}$

Calculate the value: $L = \frac{1}{(2\pi \times 855)^2 \times 1.50 \times 10^{-6}}$ $L = \frac{1}{(5370.6)^2 \times 1.50 \times 10^{-6}}$ $L = \frac{1}{288,500,000 \times 1.50 \times 10^{-6}}$ $L = \frac{1}{432.75}$ $L \approx 2.31 \times 10^{-3} \, H$ $L \approx 2.31 \, mH$

(b) To find the minimum value of the ammeter reading:

At resonance, the impedance $Z$ of the circuit is purely resistive and equals $R$.

The current $I$ can be found using Ohm's law: $I = \frac{V_{rms}}{R}$

Substitute the given values: $I = \frac{18}{1.50}$ $I = 12 \, A$

So, the answers are: (a) The inductance $L$ is approximately $2.31 \, mH$. (b) The minimum value of the ammeter reading is $12 \, A$.

To solve this problem, we need to analyze the given RLC circuit and use the information provided to find the inductance $L$ and the minimum value of the ammeter reading.

Step-by-Step Solution:

(a) Finding the Inductance $L$

1. Identify the Resonance Condition: The ammeter shows a minimum reading at the resonance frequency. At resonance, the inductive reactance $X_L$ and capacitive reactance $X_C$ are equal, and the impedance of the circuit is purely resistive.

2. Resonance Frequency Formula: The resonance frequency $f_0$ for an RLC circuit is given by: $f_0 = \frac{1}{2\pi\sqrt{LC}}$ Given $f_0 = 855 \text{ Hz}$ and $C = 1.50 \text{ MF} = 1.50 \times 10^{-6} \text{ F}$, we can solve for $L$.

3. Rearrange the Formula to Solve for $L$: $L = \frac{1}{(2\pi f_0)^2 C}$

4. Substitute the Given Values: $L = \frac{1}{(2\pi \times 855)^2 \times 1.50 \times 10^{-6}}$

5. Calculate the Inductance: $L = \frac{1}{(2\pi \times 855)^2 \times 1.50 \times 10^{-6}} \approx 2.07 \times 10^{-3} \text{ H}$ $L \approx 2.07 \text{ mH}$

(b) Finding the Minimum Value of the Ammeter Reading

1. At Resonance, the Impedance is Minimum: At resonance, the impedance $Z$ of the circuit is equal to the resistance $R$.

2. Given Resistance $R$: $R = 1.50 \text{ Qk} = 1.50 \text{ k}\Omega = 1500 \Omega$

3. Ohm's Law: The current $I$ in the circuit can be found using Ohm's Law: $I = \frac{V}{R}$ where $V$ is the RMS voltage and $R$ is the resistance.

4. Substitute the Given Values: $I = \frac{18 \text{ V}}{1500 \Omega}$

5. Calculate the Current: $I = \frac{18}{1500} \approx 0.012 \text{ A}$ $I \approx 12 \text{ mA}$

Final Answers:

(a) The inductance $L$ is approximately $2.07 \text{ mH}$.

(b) The minimum value of the ammeter reading is approximately $12 \text{ mA}$.