JEE Mains · Physics · STD 12 - 3. current electricity
The supply voltage to room is \(120\ V\). The resistance of the lead wires is \(6\,\Omega\) . A \(60\ W\) bulb is already switched on. What is the decrease of voltage across the bulb, when a \(240\ W\) heater is switched on in parallel to the bulb? ............. \(V\)
- A \(10.4 \)
- B \(0\)
- C \(2.9\)
- D \(13.3\)
Answer & Solution
Correct Answer
(A) \(10.4 \)
Step-by-step Solution
Detailed explanation
Power of bulb \(=60\, \mathrm{W}\) (given) Resistance of bulb \(=\frac{120 \times 120}{60}=240\, \Omega\left[\because \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\right]\) Power of heater \(=240\, \mathrm{W}\) (given) Resistance of heater…
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