A series R-C circuit is connected to an alter | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The current $i$ in a series $R-C$ circuit is given by:$i = \frac{V_0}{Z} = \frac{V_0}{\sqrt{R^2 + X_C^2}} = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}

Vedclass · Accepted Answer

$V_a > V_b$

Vedclass · Answer

(C) The current $i$ in a series $R-C$ circuit is given by:$i = \frac{V_0}{Z} = \frac{V_0}{\sqrt{R^2 + X_C^2}} = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}}$The voltage across the capacitor is $V = i X_C = i \cdot \frac{1}{\omega C}$.Substituting the expression for $i$:$V = \frac{V_0}{\sqrt{R^2 + (1/\omega C)^2}} \cdot \frac{1}{\omega C} = \frac{V_0}{\sqrt{R^2 \omega^2 C^2 + 1}}$When the capacitor is filled with mica (dielectric constant $K > 1$),its capacitance increases,so $C_b > C_a$.$1$. For current: As $C$ increases,$X_C = 1/\omega C$ decreases,so the impedance $Z = \sqrt{R^2 + X_C^2}$ decreases. Thus,$i_b > i_a$.$2$. For voltage across the capacitor: From $V = \frac{V_0}{\sqrt{R^2 \omega^2 C^2 + 1}}$,as $C$ increases,the denominator increases,which means $V$ decreases. Therefore,$V_b  V_b$.

$A$ series $R-C$ circuit is connected to an alternating voltage source. Consider two situations:
$(a)$ When the capacitor is air-filled.
$(b)$ When the capacitor is mica-filled.
If the current through the resistor is $i$ and the voltage across the capacitor is $V$,then:

Similar Questions

An inductor and a resistor in series are connected to an $A.C.$ supply of variable frequency. As the frequency of the source is increased,the phase angle between current and the potential difference across $L$ will

$A$ $20 V$ $AC$ is applied to a circuit consisting of a resistor and a coil with negligible resistance. If the voltage across the resistor is $12 V$,the voltage across the coil is (in $V$)

$A$ coil of $200 \, \Omega$ resistance and $1.0 \, H$ inductance is connected to an $ac$ source of frequency $200/2\pi \, Hz.$ The phase angle between potential and current will be.....$^o$

In an $LCR$ series circuit,if ${X_C} < {X_L}$ and ${X_C} > {X_L}$,determine whether the current in the circuit leads or lags the voltage.

For a series $LCR$ circuit,$R = X_L = 2X_C$. The impedance of the circuit and the phase difference between $V$ and $i$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ series $R-C$ circuit is connected to an alternating voltage source. Consider two situations:$(a)$ When the capacitor is air-filled.$(b)$ When the capacitor is mica-filled.If the current through the resistor is $i$ and the voltage across the capacitor is $V$,then: