fertilizeki

2021-12-30

Two light bulbs have constant resistances of 400 Ω and 800 Ω. If the two light bulbs are connected in series across a 120 V line, find the power dissipated in each bulb

Esta Hurtado

Beginner2021-12-31Added 39 answers

Given:

The resistance across first bulb is ${R}_{1}=400\Omega$

The resistance across second bulb is ${R}_{2}=800\Omega$

The voltage across the builbs is $V=120\text{}V$.

The formula to calculate the power dissipated in first bulb is,

$P}_{1}=\frac{{V}^{2}}{{R}_{1}$

Here, $P}_{1$ is the power dissipated in first bulb, V is the voltage and $R}_{1$ is the resistance across first bulb.

Substitute the khown values in the formula to calculate the power dissipated in first bulb.

${P}_{1}=\frac{(120V{)}^{2}}{400\Omega}\left(\frac{1W}{1{V}^{2}/\Omega}\right)$

$=36W$

The formula to calculate the power dissipated in second bulb is,

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

Here, ${P}_{2}$ is the power dissipated in first bulb and ${R}_{2}$ is the resistance across second bulb.

Substitute the known values in the formula to calculate the power dissipated in second bulb.

${P}_{2}=\frac{(120V{)}^{2}}{800\Omega}\left(\frac{1W}{1{V}^{2}/\Omega}\right)$

$=18W$

Thus, the power dissipated in first bulb is $36W$ and the power dissipated in second bulb is $18W.$

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