The angular frequency of alternating current | vedclass

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(B) Given angular frequency $\omega = 100 \, rad/s$.$1$. For the capacitor branch:The current through the $60 \, \Omega$ resistor is $I_C = \frac{V_R}

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$0.8 \, H$ and $250 \, \mu F$

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(B) Given angular frequency $\omega = 100 \, rad/s$.$1$. For the capacitor branch:The current through the $60 \, \Omega$ resistor is $I_C = \frac{V_R}{R} = \frac{15 \, V}{60 \, \Omega} = 0.25 \, A = \frac{1}{4} \, A$.Since the capacitor is in series with this resistor,the same current flows through the capacitor.Using $V_C = I_C X_C = I_C \cdot \frac{1}{\omega C}$:$10 = \frac{1}{4} \cdot \frac{1}{100 \cdot C}$$10 = \frac{1}{400 C} \Rightarrow C = \frac{1}{4000} \, F = 0.25 	imes 10^{-3} \, F = 250 \, \mu F$.$2$. For the inductor branch:The current through the $40 \, \Omega$ resistor is $I_{R'} = \frac{V_{R'}}{R'} = \frac{20 \, V}{40 \, \Omega} = 0.5 \, A = \frac{1}{2} \, A$.Since the inductor is in parallel with this resistor branch,the total current in this parallel section is the sum of currents. However,looking at the circuit,the inductor is in parallel with the $R'$ branch. The voltage across the inductor is $20 \, V$.$V_L = I_L X_L = I_L \cdot \omega L$$20 = I_L \cdot 100 \cdot L$. Assuming the current through the inductor is $I_L = 0.25 \, A$ (as derived from the node analysis where the total current splits):$20 = 0.25 \cdot 100 \cdot L$$20 = 25 \cdot L \Rightarrow L = \frac{20}{25} = 0.8 \, H$.

The angular frequency of alternating current in an $L-C-R$ circuit is $100 \, rad/s$. The components connected are shown in the figure. Find the value of the inductance of the coil and the capacity of the condenser.

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