To solve this problem, we need to determine the resistance ($R$) and inductance ($L$) of the coil, given two conditions of voltage, current, and frequency.

We are provided with:

1. When connected to a 114-volt, 60-cycle (Hz) source, the current is 3 amps.
2. When connected to a 116-volt, 25-cycle (Hz) source, the current rises to 4 amps.

Step 1: Use Ohm’s Law and Impedance Formula

The impedance of an RL circuit is given by: $Z = \sqrt{R^2 + (X_L)^2}$ where $X_L = 2\pi f L$ is the inductive reactance, and $f$ is the frequency.

From Ohm’s Law, the current $I$ can be written as: $I = \frac{V}{Z}$ where $V$ is the applied voltage, and $Z$ is the impedance of the coil.

Step 2: Set up Equations for Both Conditions

For the first condition (114V, 60Hz, 3A):

• Voltage ($V_1$) = 114V
• Frequency ($f_1$) = 60Hz
• Current ($I_1$) = 3A

From Ohm’s Law: $3 = \frac{114}{Z_1} \quad \Rightarrow \quad Z_1 = \frac{114}{3} = 38 \, \Omega$

So, the impedance at 60Hz is $Z_1 = 38 \, \Omega$.

For impedance in terms of resistance and reactance: Z_1 = \sqrt{R^2 + (X_L_1)^2} where X_L_1 = 2\pi f_1 L = 2\pi \times 60 \times L = 120\pi L.

Thus: $38 = \sqrt{R^2 + (120\pi L)^2}$

For the second condition (116V, 25Hz, 4A):

• Voltage ($V_2$) = 116V
• Frequency ($f_2$) = 25Hz
• Current ($I_2$) = 4A

From Ohm’s Law: $4 = \frac{116}{Z_2} \quad \Rightarrow \quad Z_2 = \frac{116}{4} = 29 \, \Omega$

The impedance at 25Hz is $Z_2 = 29 \, \Omega$.

For impedance in terms of resistance and reactance: Z_2 = \sqrt{R^2 + (X_L_2)^2} where X_L_2 = 2\pi f_2 L = 2\pi \times 25 \times L = 50\pi L.

Thus: $29 = \sqrt{R^2 + (50\pi L)^2}$

Step 3: Solve the System of Equations

Now we have two equations:

1. $38 = \sqrt{R^2 + (120\pi L)^2}$
2. $29 = \sqrt{R^2 + (50\pi L)^2}$

To solve these equations, we can square both sides of each equation to eliminate the square roots.

1. $38^2 = R^2 + (120\pi L)^2$
   $1444 = R^2 + (120\pi L)^2$

2. $29^2 = R^2 + (50\pi L)^2$
   $841 = R^2 + (50\pi L)^2$

Now, subtract the second equation from the first: $1444 - 841 = (120\pi L)^2 - (50\pi L)^2$ $603 = (120\pi L)^2 - (50\pi L)^2$ Factor the right-hand side: $603 = (120^2 - 50^2)\pi^2 L^2$ $603 = (14400 - 2500)\pi^2 L^2$ $603 = 11900 \pi^2 L^2$ Solve for $L^2$: $L^2 = \frac{603}{11900 \pi^2}$ $L^2 \approx \frac{603}{37219.6} \approx 0.0162$ $L \approx \sqrt{0.0162} \approx 0.127 \, \text{H}$

Step 4: Solve for $R$

Now substitute $L \approx 0.127 \, \text{H}$ back into one of the original equations to find $R$. Using the second equation: $841 = R^2 + (50\pi \times 0.127)^2$ $841 = R^2 + (19.850)^2$ $841 = R^2 + 393.4$ $R^2 = 841 - 393.4 = 447.6$ $R \approx \sqrt{447.6} \approx 21.2 \, \Omega$

Final Answer:

• Resistance, $R \approx 21.2 \, \Omega$
• Inductance, $L \approx 0.127 \, \text{H}$

Would you like any further details or clarifications? Here are some related questions you might find interesting:

1. How does the impedance of a series RL circuit change with frequency?
2. How does the current depend on the applied voltage in an RL circuit?
3. What would happen to the impedance if the frequency increased significantly?
4. How can you calculate the power factor of an RL circuit?
5. How can the phase angle between current and voltage in an RL circuit be determined?

Tip: When solving RL circuit problems, remember that impedance is frequency-dependent due to the inductive reactance, so changes in frequency affect the total impedance and current.