Question

A 100W bulb ${{B}_{1}}$ and two 60 W bulbs ${{B}_{2}}$ and ${{B}_{3}}$ are connected to a 250 V source as shown in the figure. Now, ${{W}_{1}},{{W}_{2}}\text{ and }{{W}_{3}}$ are the output powers of the bulbs ${{B}_{1}},{{B}_{2}}\text{ and }{{B}_{3}}$ respectively. Then –

& \text{A) }{{W}_{1}}>{{W}_{2}}\text{= }{{W}_{3}} \\\ & \text{B) }{{W}_{1}}>{{W}_{2}}\text{}{{W}_{3}} \\\ & \text{C) }{{W}_{1}}<{{W}_{2}}\text{= }{{W}_{3}} \\\ & \text{D) }{{W}_{1}}<{{W}_{2}}<{{W}_{3}} \\\ \end{aligned}$$

Answer 1

We need to understand the relation between the power rating of a resistor or an appliance and the voltage applied to the given circuit to understand the output powers of the devices. We can find the resistance of the device which will be constant.

Complete answer:
We are given three bulbs which have different power ratings. We are also given the voltage applied to the network as shown in the figure. We know that the current flowing in the circuit through the three bulbs are different therefore, there will be a change in the output power from the power ratings.

We know that the resistance of the bulbs will be a constant. We can find the resistance of each of the bulbs using the power rating and the applied potential difference.

& P=\dfrac{{{V}^{2}}}{R} \\\ & \therefore R=\dfrac{{{V}^{2}}}{R} \\\ \end{aligned}$$ We know that the resistance for the three bulbs can be found easily by applying this formula as – $$\begin{aligned} & {{R}_{1}}=\dfrac{{{V}^{2}}}{100} \\\ & {{R}_{2}}={{R}_{3}}=\dfrac{{{V}^{2}}}{60} \\\ \end{aligned}$$ Now, we know that the potential drop across each of the bulb 1 and bulb 2 can be given by the voltage divider rule as – $$\begin{aligned} & {{V}_{n}}=\dfrac{V{{R}_{n}}}{{{R}_{1}}+..+{{R}_{n}}} \\\ & \Rightarrow {{V}_{1}}=\dfrac{V{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \\\ & \therefore {{V}_{1}}=250\dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \\\ & \text{and,} \\\ & \Rightarrow {{V}_{2}}=\dfrac{V{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \\\ & \therefore {{V}_{2}}=250\dfrac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \\\ \end{aligned}$$ Now, we can find the output power for each of the bulbs using the relation between the voltage drop across each of the bulb and the resistance of each of the bulb as – $$\begin{aligned} & W=\dfrac{{{V}^{2}}}{R} \\\ & \Rightarrow {{W}_{1}}=\dfrac{{{V}_{1}}^{2}}{{{R}_{1}}} \\\ & \therefore {{W}_{1}}=\dfrac{{{250}^{2}}{{\left( \dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \right)}^{2}}}{{{R}_{1}}}=\dfrac{{{250}^{2}}}{{{({{R}_{1}}+{{R}_{2}})}^{2}}}{{R}_{1}} \\\ & {{W}_{2}}=\dfrac{{{V}_{2}}^{2}}{{{R}_{2}}} \\\ & \therefore {{W}_{2}}=\dfrac{{{250}^{2}}}{{{({{R}_{1}}+{{R}_{2}})}^{2}}}{{R}_{2}} \\\ & \text{and,} \\\ & \therefore {{\text{W}}_{3}}=\dfrac{{{250}^{2}}}{{{R}_{3}}} \\\ \end{aligned}$$ Now, we can find the ratio between the three output works in the situation as – $$\begin{aligned} & {{W}_{1}}:{{W}_{2}}:{{W}_{3}}=\dfrac{{{250}^{2}}}{{{({{R}_{1}}+{{R}_{2}})}^{2}}}{{R}_{1}}:\dfrac{{{250}^{2}}}{{{({{R}_{1}}+{{R}_{2}})}^{2}}}{{R}_{2}}:\dfrac{{{250}^{2}}}{{{R}_{3}}} \\\ & \Rightarrow {{W}_{1}}:{{W}_{2}}:{{W}_{3}}={{R}_{1}}:{{R}_{2}}:\dfrac{{{({{R}_{1}}+{{R}_{2}})}^{2}}}{{{R}_{3}}} \\\ & \Rightarrow {{W}_{1}}:{{W}_{2}}:{{W}_{3}}=\dfrac{{{V}^{2}}}{100}:\dfrac{{{V}^{2}}}{60}:{{\dfrac{\left( \dfrac{{{V}^{2}}}{100}+\dfrac{{{V}^{2}}}{60} \right)}{\dfrac{{{V}^{2}}}{60}}}^{2}} \\\ & \Rightarrow {{W}_{1}}:{{W}_{2}}:{{W}_{3}}=\dfrac{1500}{100}:\dfrac{1500}{60}:64 \\\ & \therefore {{W}_{1}}:{{W}_{2}}:{{W}_{3}}=15:25:64 \\\ \end{aligned}$$ The ratio between the power outputs of the three bulbs is 15:25:64. $$\therefore \text{ }{{W}_{1}}<{{W}_{2}}<{{W}_{3}}$$ **So, the correct answer is option D.** **Note:** We need to be careful while handling the potential drops across resistors as the parallel combination itself may contain a series combination of resistors as we have seen in this problem for the bulb 1 and bulb 2. We need to consider the voltage for each resistor then.