Question:
A heating element is designed to dissipate \(2400\,\text{W}\) when connected to \(240\,\text{V}\). The power it dissipates when it is connected to \(120\,\text{V}\) is \((\text{Assume that resistance of the filament is constant})\)
A heating element is designed to dissipate \(2400\,\text{W}\) when connected to \(240\,\text{V}\). The power it dissipates when it is connected to \(120\,\text{V}\) is \((\text{Assume that resistance of the filament is constant})\)
Show Hint
For a resistor with constant resistance,
\[
P=\frac{V^2}{R}.
\]
Hence, power varies as the square of the applied voltage:
\[
P\propto V^2.
\]
If the voltage is halved, the power becomes one-fourth.
Updated On: Jun 18, 2026
- \(600\,\text{W}\)
- \(1200\,\text{W}\)
- \(1800\,\text{W}\)
- \(400\,\text{W}\)
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The Correct Option is A
Solution and Explanation
Step 1: Determine the resistance of the heating element.
Given, \[ P_1=2400\,\text{W} \] and \[ V_1=240\,\text{V}. \] Using \[ P=\frac{V^2}{R}, \] we get \[ R=\frac{V_1^2}{P_1} \] \[ R=\frac{(240)^2}{2400} \] \[ R=\frac{57600}{2400} \] \[ R=24\,\Omega. \]
Step 2: Calculate the power at \(120\,\text{V}\).
Now, \[ V_2=120\,\text{V}. \] Using \[ P_2=\frac{V_2^2}{R}, \] \[ P_2=\frac{(120)^2}{24} \] \[ P_2=\frac{14400}{24} \] \[ P_2=600\,\text{W}. \]
Step 3: Alternative verification using proportionality.
Since the resistance remains constant, \[ P\propto V^2. \] Therefore, \[ \frac{P_2}{P_1} = \left(\frac{V_2}{V_1}\right)^2 \] \[ = \left(\frac{120}{240}\right)^2 \] \[ = \left(\frac12\right)^2 \] \[ = \frac14. \] Hence, \[ P_2=\frac14(2400) \] \[ P_2=600\,\text{W}. \]
Step 4: Final conclusion.
Therefore, the power dissipated at \(120\,\text{V}\) is \[ \boxed{600\,\text{W}} \]
Given, \[ P_1=2400\,\text{W} \] and \[ V_1=240\,\text{V}. \] Using \[ P=\frac{V^2}{R}, \] we get \[ R=\frac{V_1^2}{P_1} \] \[ R=\frac{(240)^2}{2400} \] \[ R=\frac{57600}{2400} \] \[ R=24\,\Omega. \]
Step 2: Calculate the power at \(120\,\text{V}\).
Now, \[ V_2=120\,\text{V}. \] Using \[ P_2=\frac{V_2^2}{R}, \] \[ P_2=\frac{(120)^2}{24} \] \[ P_2=\frac{14400}{24} \] \[ P_2=600\,\text{W}. \]
Step 3: Alternative verification using proportionality.
Since the resistance remains constant, \[ P\propto V^2. \] Therefore, \[ \frac{P_2}{P_1} = \left(\frac{V_2}{V_1}\right)^2 \] \[ = \left(\frac{120}{240}\right)^2 \] \[ = \left(\frac12\right)^2 \] \[ = \frac14. \] Hence, \[ P_2=\frac14(2400) \] \[ P_2=600\,\text{W}. \]
Step 4: Final conclusion.
Therefore, the power dissipated at \(120\,\text{V}\) is \[ \boxed{600\,\text{W}} \]
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