Question 12.42: Refer to Fig. 12.56(a), where i=Icos(ωt). Obtain an expressi...
Refer to Fig. 12.56(a), where i=I \cos (\omega t). Obtain an expression for the phasor voltage across the inductor. Check the result by obtaining expressions for the voltage for f=0 and f \rightarrow \infty.
We transform the current source and parallel resistance R_1 to a voltage source and series resistance R_1, and use voltage division to obtain
\tilde{V}=\left(\frac{R_3 \| Z_L}{R_1+R_2\left\|Z_C+R_3\right\| Z_L}\right) \tilde{I} R_1, (12.75)
where
Z_L=j \omega L, Z_C=\frac{1}{j \omega C} .
To check this result, we first examine it for f=0 and f \rightarrow \infty . For f=0, R_3 \| Z_L=0 and R_2 \| Z_C \rightarrow R_2, so \tilde{V}=0. For f \rightarrow \infty, R_3 \| Z_L \rightarrow R_3 and R_2 \| Z_C \rightarrow 0, so
\tilde{V}=\frac{R_1 R_3}{R_1+R_3} \tilde{I} .
Next we draw the circuit diagram for f=0 and for f \rightarrow \infty and obtain expressions for the voltage \tilde{V} in each case. For f=0 , the capacitor is an open circuit and the inductor is a short circuit (conductor). Because the voltage drop across a conductor equals zero, \tilde{V}=0 .
For f \rightarrow \infty, the inductor is an open circuit and the capacitor is a short circuit (conductor). The voltage \nu is the voltage across the parallel connection of R_1 and R_3, given by
\tilde{V}=\frac{R_1 R_3}{R_1+R_3} \tilde{I} .
These checks give us confidence in the expression (12.75).