Explanation
Given that,
Resistance $${{R}_{C}}=100\Omega $$
$$\omega =100\,rad/s$$
$${{X}_{L}}=50\,H$$
$${{R}_{L}}=50\Omega $$
$${{X}_{C}}=100\mu F$$
Now, the impedance is
$$ {{z}_{1}}=\sqrt{{{R}^{2}}+X_{C}^{2}} $$
$$ {{Z}_{1}}=\sqrt{{{\left( 100 \right)}^{2}}+{{\left( 100 \right)}^{2}}} $$
$$ {{Z}_{1}}=100\sqrt{2} $$
Now the current is
$${{I}_{1}}=\dfrac{20}{100\sqrt{2}}\,A$$
$$ {{Z}_{L}}=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}} \right)}^{2}}} $$
$$ {{Z}_{L}}=50\sqrt{2} $$
Now, the current is
$${{I}_{2}}=\dfrac{20}{50\sqrt{2}}\,A$$
Now, the total current is
$$ I=\sqrt{I_{1}^{2}+I_{2}^{2}} $$
$$ I=\sqrt{{{\left( \dfrac{20}{100\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{20}{50\sqrt{2}} \right)}^{2}}} $$
$$ I=\dfrac{1}{\sqrt{10}}\,A $$
Now, voltage across 100 Ω resistor is
$$ {{V}_{100}}=\dfrac{20}{100\sqrt{2}}\times 100 $$
$$ {{V}_{100}}=10\sqrt{2}\,V $$
Now, voltage across 50 Ω resistor is
$$ {{V}_{50}}=\dfrac{20}{50\sqrt{2}}\times 50 $$
$$ {{V}_{50}}=10\sqrt{2}\,V $$
Hence, the voltage across 50Ω resistor is $$10\sqrt{2}\,V$$
At resonance, the capacitive reactance and the inductive reactance of the series LCR circuit become equal to each other and cancels each other. Therefore, the total impedance of the circuit will be only due to resistance.
So __ $$R = 10\Omega $$
Emf, $$e=20\sin 300t$$
Current, $$i=4\sin 300t$$
Reactance,
$$ R=\dfrac{emf}{I} $$
$$ R = \dfrac{20}{4}=5\,ohm $$
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