First, calculate the individual resistance of the heater and the bulb:
$$R_H = \frac{V^2}{P_{H,\text{rated}}} = \frac{220^2}{750} = \frac{48400}{750} \approx 64.53\,\Omega$$
$$R_B = \frac{V^2}{P_{B,\text{rated}}} = \frac{220^2}{200} = \frac{48400}{200} = 242.0\,\Omega$$
Total resistance in series:
$$R_{\text{total}} = R_H + R_B = 64.53 + 242.0 = 306.53\,\Omega$$
Circuit current under $220\,\text{V}$ supply:
$$I = \frac{V}{R_{\text{total}}} = \frac{220}{306.53} \approx 0.7177\,\text{A}$$
Power consumed by the bulb:
$$P_B = I^2 R_B = (0.7177)^2 \times 242 \approx 0.5151 \times 242 \approx 124.8\,\text{W}$$
Power consumed by the heater:
$$P_H = I^2 R_H = (0.7177)^2 \times 64.53 \approx 0.5151 \times 64.53 \approx 33.25\,\text{W}$$
Hence, option 1 is the correct answer.
पहले हीटर और बल्ब का व्यक्तिगत प्रतिरोध ज्ञात करते हैं:
$$R_H = \frac{V^2}{P_{H,\text{rated}}} = \frac{220^2}{750} = \frac{48400}{750} \approx 64.53\,\Omega$$
$$R_B = \frac{V^2}{P_{B,\text{rated}}} = \frac{220^2}{200} = \frac{48400}{200} = 242.0\,\Omega$$
श्रेणीक्रम में कुल प्रतिरोध:
$$R_{\text{total}} = R_H + R_B = 64.53 + 242.0 = 306.53\,\Omega$$
$220\,\text{V}$ की आपूर्ति पर परिपथ में धारा:
$$I = \frac{V}{R_{\text{total}}} = \frac{220}{306.53} \approx 0.7177\,\text{A}$$
बल्ब द्वारा उपभुक्त शक्ति:
$$P_B = I^2 R_B = (0.7177)^2 \times 242 \approx 124.8\,\text{W}$$
हीटर द्वारा उपभुक्त शक्ति:
$$P_H = I^2 R_H = (0.7177)^2 \times 64.53 \approx 33.25\,\text{W}$$
अतः विकल्प 1 सही उत्तर है।
