Series Resonance Condition for Minimum Impedance

A telegraph line of length 100 km has a capacity of 0.01 μ F / km and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is mH. (if π=√(10))

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why does minimum impedance mean XL = XC?

In a series RLC circuit, impedance is Z = sqrt(R^2 + (XL - XC)^2). Here, R is fixed and non-negative. Therefore, Z is smallest when the reactive term (XL - XC)^2 is zero. This condition happens when XL = XC.

Answer

Accepted answer: 100 mH

For minimum impedance, the circuit must be in series resonance with an added inductance of $100 m H$ .

Why each option works or fails

Step-by-step solution

Given: Line length $l = 100 km$ , capacitance per unit length $c^{'} = 0.01 μ F/km = 10^{- 8} F/km$ , frequency $f = 0.5 kilo cycles/s = 500 Hz$ , and $π = \sqrt{10}$ .
Find: Value of series inductance $L$ in $mH$ required for minimum impedance.
Visualise: The telegraph line behaves as a total lumped capacitance $C = c^{'} \times l$ . Adding an inductor $L$ in series creates a series circuit whose impedance $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$ is minimized when the inductive and capacitive reactances cancel completely.
Strategise: Minimum impedance corresponds to the series resonance condition: $X_{L} = X_{C} ⟹ ω L = \frac{1}{ω C} ⟹ L = \frac{1}{ω^{2} C} = \frac{1}{4 π^{2} f^{2} C}$ .
Execute: Total capacitance is $C = 100 \times 0.01 μ F = 1 μ F = 10^{- 6} F$ . Angular frequency squared is $ω^{2} = 4 π^{2} f^{2} = 4 \times 10 \times (500)^{2} = 40 \times 250000 = 10^{7} {rad}^{2} / s^{2}$ . Therefore, $L = \frac{1}{10^{7} \times 10^{- 6}} = \frac{1}{10} H = 0.1 H = 100 mH$ .
Verify: Check dimensions: $[L] = [1 / (ω^{2} C)] = s^{2} / F = H$ . With $f = 500 Hz$ and $C = 1 μ F$ , $ω = 2 π (500) = 1000 π \approx 3162 rad/s$ , so $X_{C} = 1 / (ω C) \approx 316.2 Ω$ . Then $X_{L} = ω L = 3162 \times 0.1 = 316.2 Ω$ , perfectly canceling.

Series Resonance Condition for Minimum Impedance

What feels right?

For minimum impedance, the circuit must be in series resonance with an added inductance of $100 m H$ .

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

We think you should solve this next.

An ideal transformer with purely resistive load operates at $12 kV$ on the primary…

Two concentric circular coils with radii $1 cm$ and $1000 cm$ , and number o…

Let the eccentricity of the hyperbola…

Students also ask

Series Resonance Condition for Minimum Impedance

For minimum impedance, the circuit must be in series resonance with an added inductance of 100 mH100 mH.

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

An ideal transformer with purely resistive load operates at 12 kV12 kV on the primary…

Two concentric circular coils with radii 1 cm1 cm and 1000 cm1000 cm, and number o…

Let the eccentricity of the hyperbola…

Students also ask

For minimum impedance, the circuit must be in series resonance with an added inductance of $100 m H$ .

An ideal transformer with purely resistive load operates at $12 kV$ on the primary…

Two concentric circular coils with radii $1 cm$ and $1000 cm$ , and number o…