Question

A 1000w heating unit is designed to operate from a 200 volt.by what percentage will its heat output drop if the line voltage drops by 40 V.?

Answer 1

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Answer 2

Answer:

The heat output drops if the line voltage drops by $40V$ is $96\%$.

Explanation:

Given,

Voltage, $V_{1}$ = $200V$

The power to operate 200V voltage, $P_{1}$ = $1000W$

Voltage, $V_{2}$ = $40 V$

The heat output drops if the line voltage drops by $40V$ =?

As we know,

• According to the joule's law, power consumed is given by the ratio of the square of the voltage to the resistance.
• $P = \frac{V^{2} }{R}$

Now, when resistance is constant, then we can say;

• $\frac{P_{2} }{P_{1} } = \frac{V_{2}^{2} }{V_{1}^{2} }$

After subtracting variable 1 from both sides of the equation, we get:

• $\frac{P_{2} - P_{1} }{P_{1} } = \frac{V_{2}^{2} - V_{1}^{2} }{V_{1}^{2} }$
• $\frac{\Delta P }{P_{1} } = \frac{V_{2}^{2} - V_{1}^{2} }{V_{1}^{2} }$

Now by multiplying it by 100, we get the percentage drop:

• $\frac{\Delta P }{P_{1} } \times100= \frac{V_{2}^{2} - V_{1}^{2} }{V_{1}^{2} }\times 100$ = $\frac{(40)^{2} - (200)^{2} }{(200)^{2} }\times 100$ = $-96\%$

The negative sign determines the drop in power.

Hence, the power drop is $96\%$.