Question

A 9v battery is connected in a circuit having three resistors 3 ohm, 6ohm,9ohm connected in series. Determine the power dissipated in 6 ohm resistor

Answer 1

Given :-

•Three resistors are connected in series

3ohm,6ohm and 9ohm

• A battery of 9 volts

To find. ;-

Power (p) dissipated by 6ohm resisitor= ?

$\underline{\underline{\purple{\sf{\dag\ \ \ Solution:-}}}}$

• First find the equivalent resistance then find the current (I)

• Resistor in series are calculated by given formula !

$\boxed{\star{\sf{\ \ R_s= R_1+R_2+R_3..\ ..\ \ ..+R_n}}}$

Where

$\sf \bullet R_s=$ Equivalent resistance [sum of individual resistor's]

• Now find the value !

$:\implies\sf R_s= R_1+R_2+R_3\\ \\ :\implies\sf R_s= 3\Omega+6\Omega+9\Omega\\ \\ :\implies\sf R_s= 18\Omega$

$\boxed{\sf{\dag \ Equivalent\ resistance = 18 \: \Omega}}$

• Now calculated the current

$\bigstar{\boxed{\sf{Voltage (v)= Current (I)\times Resistance (R)}}}$

We have !

•Voltage = 9V

•Resistance = 18 ohm

•Current =?

$:\implies\sf I= \dfrac{V}{R}\\ \\ :\implies\sf I= \dfrac{9}{18}\\ \\ :\implies\sf I= \cancel{\dfrac{1}{2}} \\ \\ :\implies\sf Current (I) = 0.5\ Ampere$

$\boxed{\textsf{\textbf{Current (I) = 0.5 \ Ampere}}}$

• Finally find the power dissipated by 6 ohm resistor

$\bigstar{\boxed{\sf{\dag\ \ Power (P)= [Current]^2(I^2) \times Resistance (R)}}}$

• Put the given values to find the power !

$:\implies\sf P=I^2R\\ \\ :\implies\sf P= (0.5)^2\times 6\\ \\ :\implies\sf P= 0.25\times 6\\ \\ :\implies\sf P= 1.5 \ watts$

$\underline{\boxed{\textsf{\textbf{\dag Power \ dissipated = 1.5\ watts}}}}$

Answer 2

Answer:

1.5 watts

Explanation:

Given :

• Battery has a voltage of 9 volts
• Resistors of 3,6 and 9 ohms are connected in series

To find :

• Power dissipation in 6 ohms resistor

First we need to find the equivalent resistance in the circuit and then we need to find the power

When three resistors are connected in series, equivalent resistance = r₁ + r₂ + r₃

Where:

r₁ = 3 ohms

r₂ = 6 ohms

r₃ = 9 ohms

Equivalent resistance = 9+6+3 = 18 ohms

Voltage = 9 volts

Voltage = current×equivalent resistance

9 = I × 18

I = 9/18

I = 0.5 ampere

Power = I²R ( R value shall be 6 as we are finding power dissipated by 6 ohms resistor)

Substituting the values,

(0.5)² × 6

0.25 × 6

1.5 watts

The power dissipated by the 6 ohms resistor is 1.5 watts