Question

all of 9 please

9. (a) If current and voltage in an ac circuit depend on 1(1)10 sin(2 )and V(1)-0 sin(2m), respectively, what is the time-dependent Power? P(t)-I(t)V(t) = lovo (b) From graph, what is the time-averaged value of sin (o)? (c) What is the time-averaged power? Pave . Average power IoVo (d) How are the rms values of current and voltage defined in terms of peak values? Io ms 0 rms (e) Write the time-averaged power in terms of rms current and voltage Pave (f) What are the advantages of expressing currents and voltages in terms of rms value rather than peak values?

Answer 1

9(a) The time dependent power is,

$P(t) = I(t)V(t) \Rightarrow P(t) = I_0V_0 sin^2(2 \pi ft)$

(b) From graph we can see that the average power is half of the final value, hence we can say that the average value is,

$\left \langle P(t) \right \rangle = \frac{1}{2} I_0V_0$

$\Rightarrow \left \langle I_0 V_0 sin^2(\omega t) \right \rangle = \frac{1}{2} I_0V_0$

Since I0 and V0 are constants, we can take them out of the brackets which calculate the average.

$\Rightarrow I_0 V_0 \left \langle sin^2(\omega t) \right \rangle = \frac{1}{2} I_0V_0$

$\Rightarrow \left \langle sin^2(\omega t) \right \rangle = \frac{1}{2}$

(c) The time averaged power is,

$P_{avg} = \left \langle P(t) \right \rangle = \frac{1}{2} I_0V_0$

(d) The time averaged power can be written as,

$\Rightarrow P_{avg} = \left(\frac{I_0}{\sqrt{2}} \right ) \left(\frac{V_0}{\sqrt{2}} \right ) = I_{rms} V_{rms}$

$\Rightarrow I_{rms} = \frac{I_0}{\sqrt{2}} \,\, ; V_{rms} = \frac{V_0}{\sqrt{2}}$

(e) We can write the average power as,

$P_{avg} = I_{rms} V_{rms}$

(f) The advantage of mentioning rms values of I and V are that, it allows us to get an idea of how much power is there in the electrical energy passing through the wire. This average power can be used to design more efficient electronic circuits, as for AC signals rms values are more important than peak values.