In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :

Solution

Physics sir · 3 steps

Resonance frequency has no R in it, so raising R only flattens the peak and widens the band.

1. Resonance frequency. At resonance X_L = X_C, which gives ω₀ = 1/√LC. There is no R in this formula, so changing R does not move the resonance frequency.

   Q = (ω₀ L)/R = 1/R√(L/C), Δω = ω₀/Q = R/L

2. Effect of larger R. The quality factor Q goes as 1/R, so it decreases. The bandwidth Δω = R/L goes as R, so it increases. The current peak becomes lower and wider.
3. So. Only (a) is right: raising R increases the bandwidth, while ω₀ stays the same and Q falls.

The idea to keep. In a series LCR circuit, L and C fix where the resonance peak sits. R only decides how sharp it is: larger R means lower Q and a wider band.

Answer from the NTA final answer key. Solution by PrepWiser.