Question

Consider following series R-L-C circuit.
The maximum voltage drop across inductance is

• A. 50 volt
• B. 25 volt
• C. 12.5 volt
• D. 5 volt

Answer 1

The correct option is C 12.5 volt

$V=5\sin \left(100t\right)$

Compare witer $v=V_o\sin wt$

We get $V_o=5volt,w=100$

$\Rightarrow X_L=wL=100\times 0.5=50\surd 2$

$X_c=\frac{1}{wc}=\frac{1}{100\times 200\times {10}^{-6}}=50\surd 2$

$R=20\surd 2$

$X_L\sim X_c=0\Rightarrow$ resonance

impedance $z=\sqrt{(X_L\sim X_c{)}^2+R^2}+\sqrt{0+{20}^2}=20\surd 2$

$\Rightarrow$ max current $I=\frac{V_o}{Z}=\frac{S}{20}Amp$

$\Rightarrow$ max voltage across inductance will be

$V_2{I}_{X_L}=\frac{5}{20}\times 50\surd 2=12.5volt$

it is more than $5$ volt because its inductor