In an RLC series circuit shown in figure, the readings of voltmeters V1 and V2 are 100 V and 120 V, respectively. The source voltage is 130 V. For this situation, mark out the correct statement(s).

VR2+VL2=1002, VC−VL=120 1302=VR2+VC−VL2⇒VR=50V⇒ VL=1002−502=503V VC=120+503V cos⁡ϕ=VRV=50130=513Since VC>VL, so circuit is capacitive.

In an RLC series circuit shown in figure, the readings of voltmeters V1 and V2 are 100 V and 120 V, respectively. The source voltage is 130 V. For this situation, mark out the correct statement(s).

VR2+VL2=1002, VC−VL=120 1302=VR2+VC−VL2⇒VR=50V⇒ VL=1002−502=503V VC=120+503V cos⁡ϕ=VRV=50130=513Since VC>VL, so circuit is capacitive.

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In an RLC series circuit shown in figure, the readings of voltmeters V₁ and V₂ are 100 V and 120 V, respectively. The source voltage is 130 V. For this situation, mark out the correct statement(s).

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Power factor of the circuit is $\frac{5}{13}$

Voltage across resistor, inductor and capacitor are 50 V, $50 \sqrt{3} V$ , and $120 + 50 \sqrt{3} V$ , respectively

Voltage across resistor, inductor, and capacitor are 50 V, $50 \sqrt{3} V$ and $120 - 50 \sqrt{3} V$ , respectively

The circuit is capacitive in nature

answer is A, C, D.

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Detailed Solution

$V_{R}^{2} + V_{L}^{2} = 100^{2}, |V_{C} - V_{L}| = 120$

$\begin{array}{l} 130^{2} = V_{R}^{2} + {(V_{C} - V_{L})}^{2} \Rightarrow V_{R} = 50 V \\ \Rightarrow V_{L} = \sqrt{100^{2} - 50^{2}} = 50 \sqrt{3} V \\ V_{C} = 120 + 50 \sqrt{3} V \\ \cos ϕ = \frac{V_{R}}{V} = \frac{50}{130} = \frac{5}{13} \end{array}$