A 10-Ω resistor dissipates 1 W of power when connected to a dc voltage source. If the value of dc voltage is doubled, the resistor will dissipate
a. 1 W.
b. 2 W.
c. 4 W.
d. 10 W
The Correct Answer and Explanation is :
The correct answer is c. 4 W.
Explanation:
To solve this, we can use the relationship between power (( P )), voltage (( V )), and resistance (( R )), which is given by the formula:
[
P = \frac{V^2}{R}
]
We are told that a 10-Ω resistor dissipates 1 W of power when connected to a DC voltage source. Using this formula, we can first calculate the voltage applied across the resistor when it dissipates 1 W of power.
Step 1: Find the voltage when the power is 1 W.
We know:
- ( P = 1 \, \text{W} )
- ( R = 10 \, \Omega )
Substitute the known values into the formula:
[
1 = \frac{V^2}{10}
]
Solving for ( V^2 ):
[
V^2 = 1 \times 10 = 10
]
[
V = \sqrt{10} \approx 3.16 \, \text{V}
]
So, the initial voltage across the resistor is approximately 3.16 V.
Step 2: Double the voltage.
Now, if the voltage is doubled, the new voltage becomes:
[
V_{\text{new}} = 2 \times 3.16 = 6.32 \, \text{V}
]
Step 3: Calculate the new power dissipation.
The power dissipated by the resistor when the voltage is increased can be calculated using the same formula:
[
P_{\text{new}} = \frac{V_{\text{new}}^2}{R}
]
Substitute ( V_{\text{new}} = 6.32 \, \text{V} ) and ( R = 10 \, \Omega ):
[
P_{\text{new}} = \frac{6.32^2}{10} = \frac{39.94}{10} \approx 4 \, \text{W}
]
Thus, when the voltage is doubled, the power dissipated by the resistor increases to approximately 4 W. This result is consistent with the relationship that power dissipated in a resistor is proportional to the square of the applied voltage. Therefore, doubling the voltage causes the power to increase by a factor of four.