Question

Task 3

An alternating voltage u (V) across a resistor with resistance 10.0  varies with time

t (s) according to equation u = 25.0.sin 100t.

a) Determine the effective value of the voltage.

b) How much energy is converted into the resistor for 1 minute? EB + P / 0/0




Many years ago, in the public grid, there was an alternating voltage of the effective value 127

1.On an old lamp from that time is 127 V; 40 W. Today is the mains voltage

effective value 230

2.What would happen if this lamp were used today?

Answer 1

a) Determine the effective value of the voltage:

$$U=\frac{u}{\sqrt2}=17.7\text{ V}.$$

b) The power in the resistor is

$$P=\frac{U^2}{R}=31.3\text{ W}.$$

In 1 minute (60 seconds) it will correspond to energy of

$$E=Pt=1880\text{ J}.$$

The lamp would be much brighter. 40 W is the output power at 127 V. Therefore, the resistance is

$$R=\frac{U_1^2}{P_1}.$$

At 230 V, the power will be

$$P_2=P_1\frac{U_2^2}{U_1^2}=40\frac{230^2}{127^2}=131.2\text{ W}.$$