Question

Six similar bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance. The ratio of the power consumption by the bulbs when all the bulbs are glowing to when the two from A and one from B are glowing will be –

A) 2:1
B) 4:9
C) 9:4
D) 1:2

Answer 1

We are given a combination of bulbs connected in series and parallel to each other connected to an external source. We can find the power consumption ratios at the given conditions using the relation between the voltage and the resistance of the bulbs.

Complete step by step answer:
We know that the bulbs in a circuit constitute the loads or the resistances in the network. We can very easily approach this given case with this idea. So, six bulbs of equal resistance are connected as given below. Each bulb offers resistance of ‘R’. We need to find the equivalent resistance of the circuit including bulbs which glow in a given condition.

CASE I: All the bulbs are glowing. In this case, we can easily consider the Bulbs in section A parallel to each other and in series to the ones in B.
So, the parallel bulbs give the resistance as –

& \dfrac{1}{{{R}_{A}}}=\dfrac{1}{{{R}_{B}}}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}=\dfrac{3}{R} \\\ & {{R}_{A}}={{R}_{B}}=\dfrac{R}{3} \\\ \end{aligned}$$ Now, the equivalent resistance is given by – $${{R}_{eq}}={{R}_{A}}+{{R}_{B}}=\dfrac{2R}{3}$$ The power consumed by the network in this case is given by – $$\begin{aligned} & P=\dfrac{{{E}^{2}}}{{{R}_{eq}}} \\\ & {{P}_{1}}=\dfrac{3{{E}^{2}}}{2R}\text{ --(1)} \\\ \end{aligned}$$ CASE II: Two bulbs in section A and just one bulb in section B glows. We can simplify the circuit ass shown below.  The equivalent resistance is given by – $$\begin{aligned} & \dfrac{1}{{{R}_{A}}}=\dfrac{1}{R}+\dfrac{1}{R}=\dfrac{2}{R} \\\ & \Rightarrow {{R}_{A}}=\dfrac{R}{2} \\\ & {{R}_{B}}=R \\\ & {{R}_{eq}}={{R}_{A}}+{{R}_{B}}=\dfrac{3R}{2} \\\ \end{aligned}$$ Now, we can calculate the power dissipation by the network as – $$\begin{aligned} & P=\dfrac{{{E}^{2}}}{{{R}_{eq}}} \\\ & {{P}_{2}}=\dfrac{2{{E}^{2}}}{3R}\text{ --(2)} \\\ \end{aligned}$$ We are asked to find the ratio of power consumptions in case I and case II. This is given by taking the ratios of (1) and (2) as - $$\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{\dfrac{3{{E}^{2}}}{2R}}{\dfrac{2{{E}^{2}}}{3R}}=\dfrac{9}{4}$$ The ratio of the power consumption for the two conditions are given by 9:4. **The correct answer is option C.** **Note:** The power consumed by the electrical devices is not only dependent on its individual resistances. It is also dependent on the type of combination the circuit network is designed. This is an important point to be considered while designing a circuit.