An LCR series ac circuit is at resonance with 10 V each across $\mathrm{L}, \mathrm{C}$ and R. If the resistance is halved, the respective voltage across $\mathrm{R}, \mathrm{C}$ and L are

In an LCR series circuit at resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$, and the impedance $Z$ is equal to the resistance $R$.

Given that the voltage across each component is $V_R = V_L = V_C = 10 \text{ V}$, the current in the circuit is $I = \dfrac{V_R}{R} = \dfrac{10}{R}$.

Since $V_L = I X_L = 10 \text{ V}$ and $V_C = I X_C = 10 \text{ V}$, we have $X_L = X_C = R$.

When the resistance is halved, the new resistance becomes $R' = \dfrac{R}{2}$. The reactances $X_L$ and $X_C$ remain unchanged because the frequency of the source is constant.

The new impedance of the circuit is $Z' = \sqrt{(R')^2 + (X_L - X_C)^2} = \sqrt{\left(\dfrac{R}{2}\right)^2 + (R - R)^2} = \dfrac{R}{2}$.

The new current in the circuit is $I' = \dfrac{V}{Z'} = \dfrac{V}{R/2} = \dfrac{2V}{R} = 2I$, where $V$ is the source voltage. Since $V = V_R = 10 \text{ V}$ at resonance, the new current is $I' = \dfrac{2 \times 10}{R} = \dfrac{20}{R}$.

The new voltages are:
$V_R' = I' R' = \left(\dfrac{20}{R}\right) \left(\dfrac{R}{2}\right) = 10 \text{ V}$
$V_L' = I' X_L = \left(\dfrac{20}{R}\right) R = 20 \text{ V}$
$V_C' = I' X_C = \left(\dfrac{20}{R}\right) R = 20 \text{ V}$

Answer: $10 \mathrm{~V}, 20 \mathrm{~V}, 20 \mathrm{~V}$