A 110 V , 50 Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance 55 Ω, at resonance in the circuit, will be.... A.

Question

A 110 V , 50 Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance 55 Ω, at resonance in the circuit, will be.... A. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

At resonance I L = I C <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c73345be25b0d8cb2e76d1a305a328e4-.png" /> Alternatively, (1/ Z )=√((1/ X L )-(1/ X C ))2 At resonance, X L = X C Z arrow ∞ ∴ Z text total circuit arrow ∞ text i.e, I =0

Q. A $110 V, 50 Hz, A C$ source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω$ , at resonance in the circuit, will be.... $A$ .

At resonance $I_{L} = I_{C}$

Alternatively,
$\frac{1}{Z} = (\frac{1}{X _{L}} - \frac{1}{X _{C}})^{2}$
At resonance, $X_{L} = X_{C} & Z \to \infty$
$∴ Z_{total circuit} \to \infty i.e, I = 0$

Q. A 110V,50Hz,AC source is connected in the circuit (as shown in figure). The current through the resistance 55Ω, at resonance in the circuit, will be.... A.

At resonance IL​=IC​ Alternatively, Z1​=(XL​1​−XC​1​)2​ At resonance, XL​=XC​&Z→∞ ∴Ztotal circuit ​→∞ i.e, I=0

Q. A $110 V, 50 Hz, A C$ source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω$ , at resonance in the circuit, will be.... $A$ .

At resonance $I_{L} = I_{C}$

Alternatively,
$\frac{1}{Z} = (\frac{1}{X _{L}} - \frac{1}{X _{C}})^{2}$
At resonance, $X_{L} = X_{C} & Z \to \infty$
$∴ Z_{total circuit} \to \infty i.e, I = 0$