Question

Power dissipated in 4$\Omega$ resistor.

Answer 1

Hint : Try to find current flowing in the $4\Omega$ resistor then use the formula for power dissipated in a resistor.

Step 1 : Find equivalent resistance of the circuit.

Given: Two resistors of $8\Omega$ in parallel.

One resistance of $4\Omega$ in series with above combination.

Net resistance of two two resistors of $8\Omega$

$\frac{1}{R_{net}}=$ $\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{net}}=$ $\frac{1}{8}+\frac{1}{8}$

$⟹$ $R_{net}=4\Omega$

Equivalent resistance of whole circuit

$R_{eq}=R_{4\Omega }+R_{net}$

$⟹$ $R_{eq}=4\Omega +4\Omega =8\Omega$

Step 2 : Find current flowing in the circuit using Ohm's Law.

From Ohm's Law, $V=I{R}_{eq}$

$I=\frac{8V}{8\Omega }=1A$.

This current will flow through the $4\Omega$ resistor.

Step 3 : Find power dissipated in $4\Omega$ resistor.

Power dissipated across $4\Omega$ resistor is

$P=I^2{R}_{4\Omega }$

$P=(1A{)}^2\times 4\Omega =4W$

Final Step : The Power dissipated across the $4\Omega$ resistor is $4W$.