Question

A $60W$lightbulb and a $100W$lightbulb are placed in the circuit shown in FIGURE $EX28.9$. Both bulbs are glowing.

a. Which bulb is brighter? Or are they equally bright?

b. Calculate the power dissipated by each bulb.

Short answer

(a) $60W$bulb is brighter than$100W$bulb.

(b) $P_{60W}=23W$and $P_{100W}=14W$

Step-by-step solution

Part (a) step 1: Given information

We have given that the $60W$lightbulb and a $100W$lightbulb are placed in the circuit shown in FIGURE $EX28.9$. Both bulbs are glowing.

We need to find that which bulb is brighter or they equally bright.

Part (a) step 2: Simplify

When the current flows through a resistor, the energy is dissipated, the rate of the dissipated energy is the power. The power is given as

$P_R=\frac{{\left(∆V_R\right)}^2}{R}$

The formula ${(∆V_R)}^2/R$shows the largest dissipated power if we have several resistors. Solve equation (2) for $R$which is our target to be in the form

$R=\frac{{\left(∆V_R\right)}^2}{P_R}$

The bulbs have power $60W$and $100W$when the voltage is $120V$across each one. So, the resistance for each resistor is given as

$R_{60W}=\frac{{\left(120V\right)}^2}{60W}=240\Omega$

and

$R_{100W}=\frac{{\left(120V\right)}^2}{100W}=144\Omega$

The dissipated energy increases as the resistance increases. As $60W$bulb has the largest resistance, so it is brighter that $100W$bulb.

Part (b) step 1: Given information

We have given that $60W$lightbulb and a $100W$lightbulb are placed in the circuit shown in FIGURE $EX28.9$. Both bulbs are glowing.

We need to calculate the power dissipated by each bulb.

Part (b) step 2: Simplify

Calculating the current through the circuit by applying the loop rule by

$$\begin{gathered} V=0 \\ \in +∆V_1+∆V_2=0 \\ \in -I{R}_1-I{R}_2=0 \\ I=\frac{\in }{R_1+R_2} \end{gathered}$$

Now, putting the values for $\in$,$R_1$and $R_2$into equation (3) to get $I$

$$\begin{gathered} I=\frac{\in }{R_1+R_2} \\ =\frac{120V}{240\Omega +140\Omega } \\ =0.31A \end{gathered}$$

From Ohm's law $∆V_R=IR$, we can get an alternative formula of the dissipated power in the form

$P_R=I^{2}R$

Putting the values of$I$and $R$for each bulb

$$\begin{gathered} P_{60W}=I^2{R}_{60W}={(0.31A)}^2\left(240\Omega \right)=23W \\ {P}_{100W}=I^{2}R100W={(0.31A)}^2\left(144\Omega \right)=14W \end{gathered}$$