In a circuit,a metal filament lamp is connect | vedclass

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

Question

(A) Given: Supply voltage $V = 200 V$,Frequency $f = 50 Hz$,Power $P = 500 W$,Voltage across lamp $V_R = 100 V$.$1$. Since the circuit is an $RC$ seri

Vedclass · Accepted Answer

$100, 60$

Vedclass · Answer

(A) Given: Supply voltage $V = 200 V$,Frequency $f = 50 Hz$,Power $P = 500 W$,Voltage across lamp $V_R = 100 V$.$1$. Since the circuit is an $RC$ series circuit,the supply voltage is $V = \sqrt{V_R^2 + V_C^2}$.$200^2 = 100^2 + V_C^2 \Rightarrow V_C^2 = 40000 - 10000 = 30000$.$V_C = 100\sqrt{3} V$.$2$. The phase angle $\varphi$ is given by $	an \varphi = \frac{V_C}{V_R} = \frac{100\sqrt{3}}{100} = \sqrt{3}$.$\varphi = 	an^{-1}(\sqrt{3}) = 60^{\circ}$.$3$. Power $P = V_R \cdot I \Rightarrow 500 = 100 \cdot I \Rightarrow I = 5 A$.Also,$V_C = I \cdot X_C \Rightarrow 100\sqrt{3} = 5 \cdot X_C \Rightarrow X_C = 20\sqrt{3} \Omega$.$4$. $X_C = \frac{1}{2\pi f C} \Rightarrow 20\sqrt{3} = \frac{1}{2 \cdot \pi \cdot 50 \cdot C \cdot 10^{-6}}$.$C = \frac{1}{20\sqrt{3} \cdot 100 \cdot \pi} \cdot 10^6 = \frac{10^6}{2000 \cdot \pi \sqrt{3}} = \frac{1000}{2 \cdot 5} = 100 \mu F$.Thus,$C = 100 \mu F$ and $\varphi = 60^{\circ}$.

Similar Questions

$A$ resistance of $300 \ \Omega$ and an inductance of $\frac{1}{\pi} \ H$ are connected in series to an $AC$ voltage of $20 \ V$ and $200 \ Hz$ frequency. The phase angle between the voltage and current is

In the following circuit,an $AC$ input $V_i(t) = (20 \text{ mV}) \sin(10^5 t)$ is applied at the left end. The amplitude of the output voltage $V_0$ at the right end across the capacitor will be (in $mV$)

With an alternating voltage source of frequency $f$,an inductor $L$,a capacitor $C$,and a resistor $R$ are connected in series. The voltage leads the current by $45^{\circ}$. The value of $L$ is $(\tan 45^{\circ} = 1)$.

The time constant of the given $LR$ circuit is $\tau$. If the battery is replaced by an $AC$ source having voltage $V = V_0 \cos \omega t$,the power factor of the circuit will be

In an $LC$ circuit,the inductance $L = 40 \; mH$ and capacitance $C = 100 \; \mu F$. If a voltage $V(t) = 10 \sin (314 t)$ is applied to the circuit,the current in the circuit is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module