A 150-W lamp is placed into a 120-V ac outlet. What is the peak current?
A) 1.8 A
B) 0.80 A
C) 1.2 A
D) 0.88 A

Question

Quizplus · Accepted Answer

The peak current is given by:$$I_{peak} = \frac{V_{peak}}{Z}$$where:* $$V_{peak}$$ is the peak voltage* $$Z$$ is the impedanceThe peak voltage is given by:$$V_{peak} = V_{rms} \sqrt{2}$$where:* $$V_{rms}$$ is the root mean square voltageThe impedance is given by:$$Z = \sqrt{R^2 + X_L^2}$$where:* $$R$$ is the resistance* $$X_L$$ is the inductive reactanceThe resistance is given by:$$R = \frac{V_{rms}}{I_{rms}}$$where:* $$I_{rms}$$ is the root mean square currentThe inductive reactance is given by:$$X_L = 2\pi f L$$where:* $$f$$ is the frequency* $$L$$ is the inductanceThe frequency is given by:$$f = 60 Hz$$The inductance is given by:$$L = 0 H$$Therefore, the impedance is:$$Z = R = \frac{V_{rms}}{I_{rms}} = \frac{120 V}{1.25 A} = 96 \Omega$$Therefore, the peak current is:$$I_{peak} = \frac{V_{peak}}{Z} = \frac{120 V \sqrt{2}}{96 \Omega} = 1.8 A$$

A 150-W Lamp Is Placed into a 120-V Ac Outlet