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For the AC circuit shown below, phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$, then the values of $P$ and $Q$ are
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Verified Answer
The correct answer is:
$1 \mathrm{k} \Omega, 10 \mu \mathrm{F}$
In the shown figure, current is ahead of voltage, so its a $R C$ circuit, so $P$ is a resistor and $Q$ is a capacitor.
Now, in $R C$ circuit,
$$
Z=\sqrt{R^2+X_C^2}
$$
For simple solution,
Given
$$
\begin{aligned}
& =1000 \sqrt{2} \\
& =1000 \sqrt{1+1} \\
& =\sqrt{(1000)^2+(1000)^2} \\
\Rightarrow \quad X_C & =1000 R=1000 \Omega \\
\text { Now, } \quad X_C & =\frac{1}{\omega C} \Rightarrow C=\frac{1}{\omega X_C} \\
\Rightarrow \quad & \\
\quad \quad & =\frac{1}{100 \times 1000}=10 \mu \mathrm{F}
\end{aligned}
$$
$$
Z=1414=1000 \times 1.414
$$
Now, in $R C$ circuit,
$$
Z=\sqrt{R^2+X_C^2}
$$
For simple solution,
Given
$$
\begin{aligned}
& =1000 \sqrt{2} \\
& =1000 \sqrt{1+1} \\
& =\sqrt{(1000)^2+(1000)^2} \\
\Rightarrow \quad X_C & =1000 R=1000 \Omega \\
\text { Now, } \quad X_C & =\frac{1}{\omega C} \Rightarrow C=\frac{1}{\omega X_C} \\
\Rightarrow \quad & \\
\quad \quad & =\frac{1}{100 \times 1000}=10 \mu \mathrm{F}
\end{aligned}
$$
$$
Z=1414=1000 \times 1.414
$$
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