an-ac-power-source-with-a-variable-angular-frequency-is-connected-to-an-r-l-comb

Question

An AC power source with a variable angular frequency (ω) is connected to an R-L combination, as shown in the figure. Determine the expression of i) the current (I) in the circuit, ii) the voltage across the resistor, and iii) the voltage across the inductor as a function of ω. Hint: Use a phasor diagram.

Accepted Answer

i) I= 5.0 V / √$$(L^{2}$$ω^{2} + $$R^{2})ii) V$$_{R} = (5.0 V × R) / √$$(L^{2}$$ω^{2} + $$R^{2})$$iii) V_{L} = (5.0 V × Lω) / √$$(L^{2}$$ω^{2} + $$R^{2})$$

Answer

i) I= 5.0 V / √$$(L^{2}$$ω^{2} + $$R^{2})ii) V$$_{R} = (R) / √$$(L^{2}$$ω^{2} + $$R^{2})$$iii) V_{L} = (5.0 V × Lω) / √$$(L^{2}$$ω^{2} + $$R^{2})$$

Answer

i) I= 5.0 V / $$(L^{2}$$ω^{2} + $$R^{2})ii) V$$_{R} = (5.0 V × R) / √$$(L^{2}$$ω^{2} + $$R^{2})$$iii) V_{L} = (5.0 V × Lω) / √$$(L^{2}$$ω + $$R^{2})$$

Answer

i) I= 5.0 V / $$(L^{2}$$ω^{2} + $$R^{2})ii) V$$_{R} = (5.0 V × R) / $$(L^{2}$$ω^{2} + $$R^{2})$$iii) V_{L} = (5.0 V × Lω) / $$(L^{2}$$ω^{2} + $$R^{2})$$