An AC voltmeter connected between points A and B in the circuit b… - Vidyadip

MCQ

An $AC$ voltmeter connected between points $A$ and $B$ in the circuit below reads $36 \,V$. If it is connected between $A$ and $C$, the reading is $39 \,V$. The reading when it is connected between $B$ and $D$ is $25 \,V$. the voltmeter read when it is connected between $A$ and $D$ are ...........$V$ (Assume that, the voltmeter reads true rms voltage values and that the source generates a pure $AC$ )

✓
$\sqrt{481}$
B
$31$
C
$61$
D
$\sqrt{3361}$

✓

Answer

Correct option: A.

$\sqrt{481}$

a
(a)

Given is a series $L-C-R$ circuit.

Given,

$V_L=36 \,V \quad \dots(i)$

Given

$\sqrt{V_L^2+V_R^2}=39 \,V \quad \dots(ii)$

$\sqrt{V_C^2+V_R^2}=25 \,V \quad \dots(iii)$

So, from Eqs. $(i)$ and $(ii)$, we get

$V_L^2+V_R^2=(39)^2$

$\Rightarrow \quad V_R^2=(39)^2-(36)^2$

$\quad=(39-36)(39+36)=3(75)$

$\therefore \quad V_R=15 \,V \quad \dots(iv)$

From Eqs. $(iii)$ and $(iv)$, we have

$\quad V_C^2+(15)^2=(25)^2$

$\Rightarrow \quad V_C^2=(25)^2-(15)^2=(25-15)(25+15)$

$\therefore \quad V_C^2=10 \times 40 \Rightarrow V_C=20 V \quad \dots(iv)$

Now using,

$\quad V_{L \cdot C \cdot R}=\sqrt{V_R^2+\left(V_L-V_C\right)^2}$, we have

$V_{L-C \cdot R}=V_{A D} =\sqrt{225+(36-20)^2}$

$=\sqrt{225+256}=\sqrt{481} \,V$

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