Question 21.2: A small indicating lamp has a rating of 2 W when connected t......
A small indicating lamp has a rating of 2 W when connected to 120 V. The lamp must be connected to a voltage of 480 V at 60 Hz. A capacitor will be connected in series with the lamp to reduce the circuit current to the proper value. What value of capacitor will be needed to perform this job?
The first step is to determine the amount of current the lamp will normally draw when connected to a 120-V line:
\mathsf{I_{LAMP}}={\frac{\mathsf{p}}{\mathsf{E}}}
\mathsf{I_{LAMP}}={\frac{\mathsf{2\ W}}{\mathsf{120\ V}}}
\mathsf{I_{LAMP}} = 0.0167 A
The next step is to determine the amount of voltage that must be dropped across the capacitor when a current of 0.01667 A flows through it. Because the voltage dropped across the resistorand the voltage dropped across the capacitor are 90° out of phase with each other, vectors must be used to determine the voltage drop across the capacitor (Figure 21–12). The voltage drop across the capacitor can be calculated using the formula
\mathsf{E}_{\mathsf{C}}={\sqrt{\mathsf{E_{T}}^{2}-\mathsf{E_{R}}^{2}}}
\mathsf{E}_{\mathsf{C}}={\sqrt{\mathsf{(480\ V)}^{2}-\mathsf{(120\ V)}^{2}}}
\mathsf{E}_{\mathsf{C}}={\sqrt{216,000\ \mathsf{V}^{2}}}
\mathsf{E}_{\mathsf{C}} = 464.758 V
Now that the voltage drop across the capacitor and the amount of current flow are known, the capacitive reactance can be calculated:
{\mathsf{X_{c}}}={\frac{\mathsf{E}_{\mathsf{{C}}}}{\mathsf{I}}}
{\mathsf{X_{c}}}={\frac{\mathsf{464.758\ V}}{\mathsf{0.0167\ A}}}
{\mathsf{X_{c}}} = 27,829.82 Ω
The amount of capacitance needed to produce this capacitive reactance can now be calculated using the formula
{\mathsf{C}}=\frac{1}{2\pi {\mathsf{f X_{C}}}}
{\mathsf{C}}=\frac{1}{377\times 27,829.82\ \Omega}
C = 0.0000000953 F, or 95.3 nF
The circuit containing the lamp and capacitor is shown in Figure 21–13
