Question

A 22.10 V battery is used to supply current to a 16 k12 resistor. Assume the voltage drop across any wires used for connections is negligible. (a) What is the current (in mA) through the resistor? mA (b) What is the power dissipated by the resistor (in mW)? mW (c) What is the power input from the battery (in mW), assuming all the electrical power is dissipated by the resistor? mW (d) What happens to the energy dissipated by the resistor? The energy dissipated by the resistor is converted to elastic potential energy. The energy dissipated by the resistor is converted to gravitational potential energy. The energy dissipated by the resistor is converted to heat. The energy dissipated by the resistor is converted to electric potential energy.

Answer 1

Value of resistance given is $R=16k\Omega$

volatage of the battery is V= 22.10V

Using Ohm's law, we have V= I*R

here, V is voltage in the circuit

I is current through the resistor

R is resistance of the resistor

a) Putting the value of V and R in ohm's law equation we get

I = V/R

$I = \frac{22.10}{16*10^{3}}$

Hence,   

b) Power dissipated by the resister P is given by the formula

$P= \frac{V^{2}}{R}, P = I^{2}*R, P= V*I$

using any of these three formulas we can calculate power dissipation.

P=V*I

P= 22.10*1.38 mW

P = 30.498 mW

c) Assuming all electric power is dissipated by the resistor, Power input from the battery can be calculated by

$P= \frac{V^{2}}{R}$

Putting values of V and R ,we get

P = 30.525 mW

d) In this part the option number 3 is correct

   The energy dissipated by the resistor is converted to heat. This is because of the fact that the resistor gets hot when we supply voltage through it.