Question

Find the power dissipated in 6ohm resistance​

Answer 1

Answer:

Given:

A circuit has been provided. The voltage source is 20 V

To find:

Power dissipated in the 6 ohm resistor.

Concept:

First find out the equivalent resistance of the circuit. In that equivalent circuit , find out the current . Then the power dissipated can be easily calculated.

Power is the amount of energy dissipated by the resistance per unit time.

Calculation:

See the diagram for simplified circuit.

The 5 ohm, 4 ohm and 6 ohm are in series.

So net current will be :

$I = \dfrac{V}{(5 + 4 + 6)}$

$= > I = \dfrac{20}{(15)}$

$= > I = \dfrac{4}{3} amp.$

Power dissipated in the 6 ohm resistor will be :

$P = {(I)}^{2} \times 6$

$= > P = {( \dfrac{4}{3} )}^{2} \times 6$

$= > P = \dfrac{16}{9} \times 6$

$= > P = 10.66 \: watt$

So final answer :

$\huge{ \green{ \sf{ \bold{ P = 10.66 \: watt}}}}$

Answer 2

$\underline{\boxed{ \huge\purple{ \rm{Answer}}}}$

First see the attached image for better understanding....

Given :

voltage of battery source = 20V

To Find :

Power dissipated in 6 ohm resistor

Concept :

First we have to calculate current flow in curcuit after that we can calculate power dissipated in resistor.

Formula :

As per ohm's law V = IR

For power dissipated in resistor...

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \dag \: \boxed{ \pink{ \bold{ \rm{ P = {I}^{2} R}}}} \: \dag$

Calculation :

$\implies \rm \: eq.resistance \: in \: series \: connection \\ \\ \implies \rm \: R{ \tiny{eq}} = 6 + 9 = 15 \: \Omega \\ \\ \implies \rm \: \red{I = \frac{V}{R{ \tiny{eq}}} = \frac{20}{15} = \frac{4}{3} \: A } \\ \\ \implies \rm \: power \: dissipated \: in \: 6 \: \Omega \: resistor... \\ \\ \implies \rm \: P = {I}^{2} R = { (\frac{4}{3} )}^{2} \times 6 \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{P = 10.67 \: watt}}}}} \: \purple{ \clubsuit}$