# Power dissipation calculation across resistor from FM equation

## Calculate the dissipation in power across 20Ω resistor for the FM signal

v(t)= 20 cos(6600t+ 10sin2100t)

**Options:**

a) 5W

b) 20W

c) 10W

d) 400W

Correct Answer: a) 5W

**Explanation:**

A standard FM signal is represented by

v(t)= A_{c }cos(2πf_{c}t+ k_{f}sin2πf_{m}t)

A_{c} = carrier amplitude

f_{c}= carrier frequency

k_{f}= modulation index

f_{m }= modulating frequency

k_{f }= frequency deviation/modulating frequency

the power dissipated across 20Ω resistor is given by

V_{rms}^{2}/R

=(20/√2)^{2}/R

= 5W

#### Discussion

**RE: Power dissipation calculation across resistor from FM equation -Talha (08/27/19)**- 10 is correct ans
**RE: Power dissipation calculation across resistor from FM equation -Hansen (04/10/17)**- Shouldn't it be 10W? (20/sqrt(2))^2/20?