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Question: When the rms voltages \({{V}_{L}},{{V}_{c}}\) and \({{V}_{R}}\) are measured across the inductor L, ...

When the rms voltages VL,Vc{{V}_{L}},{{V}_{c}} and VR{{V}_{R}} are measured across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio VL:VC:VR=1:2:3{{V}_{L}}:{{V}_{C}}:{{V}_{R}}=1:2:3. If the rms voltage of the AC source is 100V, then VR{{V}_{R}} is close to:
(A) 50V
(B) 70V
(C ) 90V
(D) 100V

Explanation

Solution

Given that the VL:VC:VR=1:2:3{{V}_{L}}:{{V}_{C}}:{{V}_{R}}=1:2:3. Hence, the rms voltages VL,Vc{{V}_{L}},{{V}_{c}} and VR{{V}_{R}} are in the can be denoted as K, 2K and 3K where, K is a constant. Then use the equation for the rms voltage of the AC source. By substituting the values of VL,Vc{{V}_{L}},{{V}_{c}} and VR{{V}_{R}} and rearranging the equation we will get the value of K. Then we will get the value of the rms voltage VR{{V}_{R}} .

Complete step by step answer:
Given that,
VL:VC:VR=1:2:3{{V}_{L}}:{{V}_{C}}:{{V}_{R}}=1:2:3
V=100 volts
Since the three values are here in ratios let K be a constant. Then,
⇒VR=3K VC=2K VL=K \begin{aligned} & \Rightarrow {{V}_{R}}=3K \\\ & {{V}_{C}}=2K \\\ & {{V}_{L}}=K \\\ \end{aligned}
Then use the equation for the rms voltage of the AC source. That is,
V=VR2+(VL−VC)2V=\sqrt{{{V}_{R}}^{2}+{{\left( {{V}_{L}}-{{V}_{C}} \right)}^{2}}}
Substituting the values of VL,Vc{{V}_{L}},{{V}_{c}} and VR{{V}_{R}} and solving we get,
100=(3K)2+(K−(2K))2 ⇒100=9K2+(−K)2 ⇒100=9K2+K2 \begin{aligned} & 100=\sqrt{{{\left( 3K \right)}^{2}}+{{\left( K-\left( 2K \right) \right)}^{2}}} \\\ & \Rightarrow 100=\sqrt{9{{K}^{2}}+{{\left( -K \right)}^{2}}} \\\ & \Rightarrow 100=\sqrt{9{{K}^{2}}+{{K}^{2}}} \\\ \end{aligned}
⇒100=10K2\Rightarrow 100=\sqrt{10{{K}^{2}}}
⇒100=10K ⇒K=10010=1010 \begin{aligned} & \Rightarrow 100=\sqrt{10}K \\\ & \Rightarrow K=\dfrac{100}{\sqrt{10}}=10\sqrt{10} \\\ \end{aligned}
Now substitute the value of K in the equation of VR{{V}_{R}}. Then,
VR=3K ⇒VR=3×1010 ∴VR=3010=94.86volts \begin{aligned} & {{V}_{R}}=3K \\\ & \Rightarrow {{V}_{R}}=3\times 10\sqrt{10} \\\ & \therefore {{V}_{R}}=30\sqrt{10}=94.86volts \\\ \end{aligned}
Here since it is a close value to 94.86 volts, VR≃90volts{{V}_{R}}\simeq 90volts

Hence, option (C ) is correct.

Additional information:
When a value is given for ac voltage or current, it is ordinary the rms value. The voltage across the terminals of an outlet in room temperature is normally 240V.
The power rating of an element used in ac circuits is the average power rating. The power consumed in a circuit is always positive.

Note:
An alternating current (AC) always reverses direction and changes the magnitude direction continuously. Whereas the direct current, doesn’t reverse its polarity. Both the alternating current and direct current are measured in amperes. The rms voltage is the AC waveform of DC equivalent circuit. and the highest voltage is called the peak value.