Question

In the given figure, all bulbs are identical. The bulb that lights most brilliantly is:

A) 1 only
B) 2 only
C) 2 and 4
D) 1 and 5

Answer 1

In order to solve this problem, we have to understand the trends of the electrical properties such as voltage and current when there is a combination of two or more resistances in either series or parallel. This can help us understand the fusing of the bulb that can cause all the bulbs in the circuit to switch off.

Complete solution:
When there are two or more resistances to be connected in the circuit, the connections can be performed in two different types:
i) Series
ii) Parallel
In the series, the end of one resistor is connected to the start of another resistor and all the other resistors are connected end to end, like a chain.
In the parallel, the ends of all the resistors are bundled and connected to one terminal.
The net resistance in a series circuit, is given by:
$R = {R_1} + {R_2} + .. + {R_n}$
The net resistance in the parallel circuit is given by:
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + .. + \dfrac{1}{{{R_n}}}$
Now, let us study the variation of the current in the series and parallel circuits.
When two or more resistors are connected in series, the current flowing through all the resistors will be the same.
When two or more resistors are connected in parallel, the main current splits into the branches and the branch current flows through the individual resistances which depends on the value of the individual resistances.
If ${R_i}$ is the resistance of a resistor in parallel and V is the voltage across the branch, the branch current is given by –
${I_i} = \dfrac{V}{{{R_i}}}$
Let us consider the following circuit:

In this circuit, the bulb 1 is in series with the bulb 5 and the parallel combination of the bulbs 2, 3 and 4.
Hence, the main current will be the same through bulb 1, bulb 5 and the entire parallel combination of the bulbs 2, 3 and 4.
Thus, we can say that the main current will be split into 3 ways to light the bulbs 2,3 and 4. Thus, the current in the bulbs 2,3 and 4 will be lesser than that of 1 and 5.
Hence, bulbs 1 and 5 glow brighter than the bulbs 2, 3 and 4.

Hence, the correct option is Option D.

Note: Since there is reduction in the glow in a set of bulbs in parallel compared to others, the bulbs in the serial set lighting are never connected in parallel. So, in serial set lighting, the individual bulbs are connected in series so that all the bulbs have the same intensity of glow.