Question

The impedance of a coil connected across a single-phase AC source is \( 3 + j4 \) ohms. If another capacitive load with an impedance of \( 3 - j4 \) ohms is connected in parallel with this coil, the overall impedance seen by the AC source will be

• A. 25 ohms
• B. \( \frac{6}{25} \) ohms
• C. \( \frac{25}{6} \) ohms
• D. 6 ohms

Hint:

For parallel impedance, use the formula \( \frac{1}{Z_{\text{total}}} = \frac{1}{Z_1} + \frac{1}{Z_2} \) and simplify the complex terms.

Answer 1

The Correct Option is D

Step 1: Formula for parallel impedance.
The total impedance \( Z_{\text{total}} \) for two components in parallel is given by: \[ \frac{1}{Z_{\text{total}}} = \frac{1}{Z_1} + \frac{1}{Z_2}. \]
Step 2: Substitute the given impedances.
Given that \( Z_1 = 3 + j4 \) ohms and \( Z_2 = 3 - j4 \) ohms, we substitute these into the formula: \[ \frac{1}{Z_{\text{total}}} = \frac{1}{3 + j4} + \frac{1}{3 - j4}. \] Simplifying: \[ \frac{1}{Z_{\text{total}}} = \frac{(3 - j4) + (3 + j4)}{(3 + j4)(3 - j4)} = \frac{6}{9 + 16} = \frac{6}{25}. \] Thus: \[ Z_{\text{total}} = 6 \, \text{ohms}. \]
Step 3: Conclusion.
Thus, the overall impedance is 6 ohms, corresponding to option (D).

Answer 2

This question asks for the combined impedance of a coil \( Z_1 = 3+j4 \) ohms placed alongside a capacitive load \( Z_2 = 3-j4 \) ohms — note that \( Z_2 \) is exactly the complex conjugate of \( Z_1 \). A key property of two conjugate impedances is that their reactive (imaginary) parts are equal in magnitude but opposite in sign, so combining them cancels the reactive part completely, leaving only a real-valued resistive term. Let's use this property to check each option.

1. 25 ohms: This is the value of the product \( Z_1 Z_2 = 3^2+4^2 = 25 \), a quantity that appears in the working but is not itself the final combined impedance.
2. \( \frac{6}{25} \) ohms: This numeric value corresponds to a reciprocal (admittance-style) quantity, with units of siemens rather than ohms, so it cannot be the impedance itself.
3. \( \frac{25}{6} \) ohms: This is the value obtained by inverting the admittance-style quantity above; it does not reflect the direct cancellation of the reactive components.
4. 6 ohms: Adding the two conjugate impedances directly, \( Z_1+Z_2 = (3+j4)+(3-j4) = 6 + j0 \): the \( +j4 \) and \( -j4 \) reactive terms cancel exactly, leaving a purely real, resistive value of 6 ohms as the overall impedance.

Because the two impedances are complex conjugates, their reactive parts cancel completely when combined, leaving a purely resistive overall impedance of 6 ohms.

Therefore, the correct answer is 6 ohms.