Question

Draw a schematic diagram of a circuit consisting of a battery of three cells of $2V$ each, a $5\Omega$ resistor, an $8\Omega$ resistor, a $12\Omega$ resistor, and a plug key, all connected in series. Also, calculate the value of current through the circuit.

Answer 1

Step 1: Given data

1. Three Cells of 2V.
2. 3 resistors of $5\Omega$, $8\Omega$, and $12\Omega$.
3. A plug key

Step 2: Electric Circuit, Battery, Cell, Resistor, plug key, and Series connection:

1. Electric Circuit- An electric circuit consists of a device that provides energy to the charged particles that provide the current, such as a battery or a generator, devices that use current, and the wires or transmission lines that connect them.
2. Cell- A cell is a device that uses an electrochemical oxidation-reduction cycle to convert chemical energy stored in its active components directly into electric energy
3. Battery- A combination of cells is called a battery.
4. Resistor In an electronic circuit, a resistor is an electrical component that controls or regulates the passage of electrical current.
5. Plug key- When pressed, the plug key acts as a switch, allowing or interrupting current flow.
6. Series combination- When the same amount of current passes through the resistors, the circuit is said to be linked in series.

The total resistance in a series circuit, $R_{Total}=R_1+R_2+.....+R_n$

Where $R_{Total}$ is the total resistance of the circuit, $R_1,R_2,R_n$ are the resistance of the resistor connected.

According to Ohm's law,

$I=\frac{V}{R_{Total}}$

where $R_{Total}$ is the total resistance in the circuit, $V$ is the potential difference across the circuit, and $I$ is the total current in the circuit.

The voltage of the battery = $2V+2V+2V=6V$

Step 3: Circuit diagram,

Step 4: Current in the circuit:

The total resistance of the circuit, $R_{Total}=R_1+R_2+R_3$

Where $R_1=5\Omega ,R_2=8\Omega ,R_3=12\Omega$

$$\begin{gathered} R_{Total}=5\Omega +8\Omega +12\Omega \\ {R}_{Total}=25\Omega \end{gathered}$$

The voltage across the circuit, $V=6volts$

Current in circuit, $I=\frac{V}{R_{Total}}$

$$\begin{gathered} \Rightarrow I=\frac{6V}{25\Omega } \\ \Rightarrow I=0.24A \end{gathered}$$

Thus, the current through the circuit is $0.24A$.