An alternating current of 1.5 mA and angular frequency 300 rad / sec flows through a 10 k Ω resistor and a 0.50 μ F capacitor in series. Find the rms voltage across the capacitor and impedance of the circuit.

Question

An alternating current of 1.5 mA and angular frequency 300 rad / sec flows through a 10 k Ω resistor and a 0.50 μ F capacitor in series. Find the rms voltage across the capacitor and impedance of the circuit. (A) 10 V ; 1.2 × 104 Ω (B) 104 V ; 104 Ω (C) 50 V ; 1.0 × 104 Ω (D) 10 V ; 10 Ω

Tardigrade Admin · Accepted Answer

Iv=1.5 mA =1.5 × 10-3 A ω=300 rad / sec , R =10 k Ω=104 Ω C=0.50 μ F=0.50 × 10-6 F XC=(1/ω C)=(1/300 × 0.5 × 10-6) =6.67 × 103 Ω Z=√R2+Xc2 =√(104)2+(6.67 × 103)2 Z=1.2 × 104 Ω RMS voltage across the capacitor =Iv Xc =1.5 × 10-3 × 6.67 × 103=10 V

Q. An alternating current of $1.5 m A$ and angular frequency $300 r a d / sec$ flows through a $10 k Ω$ resistor and a $0.50 μ F$ capacitor in series. Find the rms voltage across the capacitor and impedance of the circuit.

$10 V; 1.2 \times 1 0^{4} Ω$

$1 0^{4} V; 1 0^{4} Ω$

$50 V; 1.0 \times 1 0^{4} Ω$

$10 V; 10Ω$

Q. An alternating current of 1.5mA and angular frequency 300rad/sec flows through a 10kΩ resistor and a 0.50μF capacitor in series. Find the rms voltage across the capacitor and impedance of the circuit.

10V;1.2×104Ω

104V;104Ω

50V;1.0×104Ω

10V;10Ω

Iv​=1.5mA=1.5×10−3A ω=300rad/sec,R=10kΩ=104Ω C=0.50μF=0.50×10−6F XC​=ωC1​=300×0.5×10−61​ =6.67×103Ω Z=R2+Xc2​​ =(104)2+(6.67×103)2​ Z=1.2×104Ω RMS voltage across the capacitor =Iv​Xc​ =1.5×10−3×6.67×103=10V

Q. An alternating current of $1.5 m A$ and angular frequency $300 r a d / sec$ flows through a $10 k Ω$ resistor and a $0.50 μ F$ capacitor in series. Find the rms voltage across the capacitor and impedance of the circuit.

$10 V; 1.2 \times 1 0^{4} Ω$

$1 0^{4} V; 1 0^{4} Ω$

$50 V; 1.0 \times 1 0^{4} Ω$

$10 V; 10Ω$