Question

In which circuit will the bulb or bulbs glow the brightest?
(A) A simple circuit with one bulb and one cell
(B) A simple circuit with one bulb and two cells
(C) A simple circuit with two bulbs and one cell
(D) A simple circuit with two bulbs and two cells

Answer 1

In this question we have to find the circuit in which the bulb or bulbs will glow the brightest. We know that the circuit which has the maximum power output would have the bulb which will glow the brightest. So, the power of the circuit depends upon the voltage, current and the resistance of the circuit.

Complete step by step answer:
Given:
The formula for the power output $\left( P \right)$ of the circuit having voltage $\left( V \right)$ , current $\left( I \right)$ and resistance $\left( R \right)$ is given by –
$P = V \times I$
Now according to the Ohm’s Law, the relation between voltage $\left( V \right)$ , current $\left( I \right)$ and resistance $\left( R \right)$ is given by –
$I = \dfrac{V}{R}$
Substituting the value of current $\left( I \right)$ in the power formula we get,
$\begin{array}{c} P = V \times \dfrac{V}{R}\\\ \Rightarrow P = \dfrac{{{V^2}}}{R} \end{array}$
In the given simple circuit, the bulb represents the load or the resistance $\left( R \right)$ and the cell represents the voltage $\left( V \right)$ supplied.

Now let us consider each option one by one and calculate the power output for each circuit.

Consider Option (A), there is one bulb and one cell in this circuit.
So, the voltage supplied $= V$
And, the resistance $= R$
Then, the power of this circuit
$P = \dfrac{{{V^2}}}{R}$

Similarly, Option (B), there is one bulb and two cells in this circuit.
So, the voltage supplied $= 2V$
And, the resistance $= R$
Then, the power of this circuit
$\begin{array}{c} P = \dfrac{{{{\left( {2V} \right)}^2}}}{R}\\\ \Rightarrow P = \dfrac{{4{V^2}}}{R} \end{array}$

Option (C):
There are two bulbs and one cell in this circuit.
So, the voltage supplied $= V$
And, the resistance $= 2R$
Then, the power of this circuit
$P = \dfrac{{{V^2}}}{{2R}}$

Option (D), there are two bulbs and two cells in this circuit.
So, the voltage supplied $= 2V$
And, the resistance $= 2R$
Then, the power of this circuit
$\begin{array}{c} P = \dfrac{{{{\left( {2V} \right)}^2}}}{{2R}}\\\ \Rightarrow P = \dfrac{{4{V^2}}}{{2R}}\\\ \Rightarrow P = \dfrac{{2{V^2}}}{R} \end{array}$
Now comparing the power output of each circuit, we see that in (B) option the value of the power $\left( P \right)$ is highest.
Therefore, the bulb will glow the brightest in option (B)

So, the correct answer is “Option B”.

Note:
It should be noted that for a simple circuit the power output is directly proportional to the square of the voltage supplied in the circuit and also the power output is inversely proportional to the resistance of the circuit.