A 10 ohm resistor dissipates 1W of power when connected to a DC voltage source if the DC voltage is double the resistor will dissipate how much power
The Correct Answer and Explanation is :
To solve this problem, we need to apply the relationship between power, resistance, and voltage. The formula for power dissipation in a resistor is:
[
P = \frac{V^2}{R}
]
Where:
- ( P ) is the power dissipated in watts (W),
- ( V ) is the voltage across the resistor (in volts),
- ( R ) is the resistance (in ohms).
Step-by-step Calculation:
- Given information:
- Resistance ( R = 10 \, \Omega ),
- Power dissipated ( P = 1 \, W ).
- Use the formula to find the initial voltage: We are told that the resistor dissipates 1W when connected to a DC voltage source, so we can rearrange the power formula to solve for ( V ): [
P = \frac{V^2}{R} \quad \Rightarrow \quad V^2 = P \times R
] Substituting the given values for power and resistance: [
V^2 = 1 \, \text{W} \times 10 \, \Omega = 10 \quad \Rightarrow \quad V = \sqrt{10} \approx 3.16 \, \text{V}
] So the initial voltage across the resistor is approximately 3.16V. - Now, we double the voltage: If the voltage is doubled, the new voltage ( V_{\text{new}} ) becomes: [
V_{\text{new}} = 2 \times 3.16 \, \text{V} = 6.32 \, \text{V}
] - Calculate the new power dissipation: Using the same power formula, we can now find the power dissipated when the voltage is doubled: [
P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{(6.32)^2}{10} = \frac{39.94}{10} = 3.994 \, \text{W}
] So, when the voltage is doubled, the power dissipated by the resistor is approximately 4 watts.
Explanation:
The power dissipated in a resistor is directly related to the square of the voltage applied across it. If the voltage is doubled, the power increases by a factor of four, because:
[
P \propto V^2
]
Thus, doubling the voltage results in four times the original power dissipation. In this case, since the initial power was 1W, doubling the voltage results in approximately 4W of power dissipation.