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Resistance of the bulb: $R_b = \frac{V^2}{P} = \frac{(120)^2}{60} = 240\,\Omega$
Before heater is switched on: Current $= \frac{120}{240+6} = \frac{120}{246}\,\text{A}$. Voltage across bulb $= \frac{120 \times 240}{246} \approx 116.9\,\text{V}$.
Resistance of heater: $R_h = \frac{(120)^2}{240} = 60\,\Omega$. Parallel combination: $R_p = \frac{240 \times 60}{300} = 48\,\Omega$.
After heater is switched on: Voltage across combination $= \frac{120 \times 48}{48+6} = \frac{5760}{54} \approx 106.67\,\text{V}$.
Decrease $= 116.9 - 106.67 \approx 10.04\,\text{V}$. Answer: D.
Let $n$ be a positive integer divisible by 6. Compare:
Quantity A: The remainder when $n$ is divided by 12
Quantity B: The remainder when $n$ is divided by 18
Since $n$ is divisible by 6, $n$ can be $6, 12, 18, 24, 30, 36, \ldots$
Example 1: $n = 12$. Remainder รท12 $= 0$. Remainder รท18 $= 12$. B > A.
Example 2: $n = 18$. Remainder รท12 $= 6$. Remainder รท18 $= 0$. A > B.
Example 3: $n = 36$. Remainder รท12 $= 0$. Remainder รท18 $= 0$. Equal.
The relationship changes, so it cannot be determined.
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