Question

What is the value of $R_C$ in the circuit at resonance? 

• A. $4\,\Omega$
• B. $5\,\Omega$
• C. $6\,\Omega$
• D. $7\,\Omega$

Hint:

In parallel resonance problems, always work with admittance instead of impedance. Resonance occurs when the net susceptance (imaginary part of admittance) becomes zero.

Answer 1

The Correct Option is A

Step 1: Write the admittance of the R-L branch: \(Y_1 = \dfrac{1}{5+j5} = 0.1-j0.1\).
Step 2: Write the admittance of the R_C-C branch: \(Y_2 = \dfrac{1}{R_C-j2} = \dfrac{R_C+j2}{R_C^2+4}\).
Step 3: At resonance the total susceptance is zero, so \(\dfrac{2}{R_C^2+4} = 0.1\), giving \(R_C^2 = 16\).
\[ \boxed{R_C = 4\,\Omega} \]

Answer 2

Resonance can also be confirmed by combining the two branch impedances directly into a single equivalent impedance and checking that its imaginary part vanishes, rather than working with admittances and susceptances.

The two branches are \(Z_1 = 5+j5\,\Omega\) and \(Z_2 = R_C - j2\,\Omega\), and their combined parallel impedance is \(Z_{eq} = \dfrac{Z_1 Z_2}{Z_1+Z_2}\). At resonance, \(Z_{eq}\) must come out purely real.

1. \(4\,\Omega\): \(Z_2 = 4-j2\), \(Z_1 Z_2 = 30+j10\), \(Z_1+Z_2 = 9+j3\). Dividing, \(Z_{eq} = \dfrac{30+j10}{9+j3} = 3.33 + j0\,\Omega\), which is purely real, confirming resonance.
2. \(5\,\Omega\): \(Z_2=5-j2\), \(Z_1Z_2=35+j15\), \(Z_1+Z_2=10+j3\). This gives \(Z_{eq}\) with a nonzero imaginary part, so this value does not satisfy resonance.
3. \(6\,\Omega\): \(Z_2=6-j2\), \(Z_1Z_2=40+j20\), \(Z_1+Z_2=11+j3\), giving \(Z_{eq} \approx 3.85+j0.77\,\Omega\), which still has a nonzero imaginary part, so resonance is not achieved at this value either.
4. \(7\,\Omega\): Carrying through the same combination gives an even larger residual imaginary part, so this value is further from resonance.

Only \(R_C=4\,\Omega\) reduces the combined branch impedance to a purely real number, which is exactly the condition for resonance.

Therefore, the correct answer is \(4\,\Omega\).