Chapter 19, Problem 1OQ

To determine

The battery connection that produces more light.

Answer to Problem 1OQ

Solution: b. Circuit 2 will produce more light.

Explanation of Solution

Given:

All the elements are identical in both the circuits.

Circuit diagram

  

  Figure 1

Formula used:

Ohm’s law is given as

  $I=\frac{V}{R}$

The power of the circuit is given as

  $P=VI$

Here,

$I$ is current in the circuit

  $V$ is voltage

  $R$ is the resistance

  $P$ is power

Calculation:

In the circuit 1, the bulbs (resistance) are attached in series connection and the total resistance of series connection is given as:

  $\begin{array}{c}{R}_T=R+R\\ =2R\end{array}$

Therefore, the current in the circuit is calculated as,

  $I=\frac{V}{2R}=\frac{0.5V}{R}$

And the power (light) of the circuit 1 is calculated as:

  $P_1=V\times \frac{0.5V}{R}=\frac{0.5V^2}{R}$ ...... (1)

In the circuit 2, the bulbs (resistance) are attached in parallel connection and the total resistance of parallel connection is given as:

  $\begin{array}{c}\frac{1}{R_T}=\frac{1}{R}+\frac{1}{R}\\ R_T=0.5R\end{array}$

Therefore, the current in the circuit is calculated as:

  $I=\frac{V}{R/2}=\frac{2V}{R}$

And the power (light) of the circuit 1 is calculated as:

  $P_2=V\times \frac{2V}{R}=\frac{2V^2}{R}$ ...... (2)

From the Eq. (1) and Eq. (2), the circuit 2 will produce more light.

Conclusion:

Parallel circuit will produce more light, if there are same elements.