Electricity: Circuits and their Components - Electric cell

A toy car requires a $9\text{ V}$ power supply. If you are using electric cells that provide $1.5\text{ V}$ each, how many cells must be connected in series to form the battery?

Number of cells $= \frac{9\text{ V}}{1.5\text{ V}} = 6$

Since the total voltage $V_{\text{total}}$ in a series circuit is the sum of individual voltages, we divide the required total voltage by the voltage of a single cell: $n = \frac{V_{\text{total}}}{V_{\text{cell}}}$. Thus, $6$ cells are needed.

Calculate the total voltage of a battery formed by connecting four cells in series, where each cell has a potential of $2\text{ V}$.

$V_{\text{total}} = 2\text{ V} + 2\text{ V} + 2\text{ V} + 2\text{ V} = 8\text{ V}$

Using the formula for cells in series: $V_{\text{total}} = V_1 + V_2 + V_3 + V_4$. Adding the values: $2 + 2 + 2 + 2 = 8\text{ V}$.

A student connects three cells of $1.5\text{ V}$ each. Calculate the total voltage using vertical addition.

$\begin{array}{r} 1.5 \\ 1.5 \\ + 1.5 \\ \hline 4.5 \end{array}$

The total voltage is the sum of the individual voltages of the three cells: $1.5 + 1.5 + 1.5 = 4.5\text{ V}$.

A flashlight uses five $1.2\text{ V}$ rechargeable cells connected in series. Using vertical addition, calculate the total voltage supplied to the flashlight bulb.

$\begin{array}{r} 1.2 \\ 1.2 \\ 1.2 \\ 1.2 \\ + 1.2 \\ \hline 6.0 \end{array}$ Total Voltage = $6.0\text{ V}$

In a series combination, the total voltage is the sum of the voltages of all individual cells. Adding $1.2$ five times gives $6.0$.

A scientific equipment requires a total voltage of $4.5\text{ V}$. If you are using identical electric cells each providing $1.5\text{ V}$, how many cells need to be connected in series? Represent the final circuit with a battery, a switch, and a resistor.

$\text{Number of cells} = \frac{\text{Total Voltage}}{\text{Voltage per cell}}$ $\text{Number of cells} = \frac{4.5\text{ V}}{1.5\text{ V}}$ $\text{Number of cells} = 3$

$\begin{array}{r} 1.5 \\ 1.5 \\ + 1.5 \\ \hline 4.5 \end{array}$

To achieve a higher voltage, cells are connected in series where the positive terminal of one cell is connected to the negative terminal of the next. By dividing the required total voltage ($4.5\text{ V}$) by the voltage of a single cell ($1.5\text{ V}$), we find that 3 cells are needed.