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Step-by-Step Solution
Step 1: Determine the resistance of the lamp
The lamp requires 10 A at 80 V in DC mode. Its effective resistance, R, can be found by using Ohm’s law:
$R = \frac{80}{10} = 8\,\Omega$
Step 2: Apply the AC condition
When connected to a 220 V (rms), 50 Hz AC source, the lamp’s total impedance (due to the resistor and inductor in series) must draw 10 A. Thus, we use the relation for current in an AC circuit:
$10 = \frac{220}{\sqrt{R^2 + X_L^2}}$
where $X_L$ is the inductive reactance.
Step 3: Solve for inductive reactance
Rearrange to get an expression for $X_L$:
$\sqrt{R^2 + X_L^2} = 22 \quad \Rightarrow \quad R^2 + X_L^2 = 22^2 = 484$
$X_L^2 = 484 - 64 = 420 \quad \Rightarrow \quad X_L = \sqrt{420} \approx 20.5\,\Omega$
Step 4: Relate inductive reactance to inductance
The inductive reactance $X_L$ is given by:
$X_L = \omega L \quad \text{where} \quad \omega = 2\pi f$
For a frequency $f = 50\,\text{Hz}$, we have
$\omega = 2 \pi \times 50 = 314\,\text{rad/s}$
So,
$L = \frac{X_L}{\omega} = \frac{20.5}{314} \approx 0.065\,H$
Step 5: Conclusion
The required inductor is approximately 0.065 H.
