Question

Question $1$: Draw a schematic diagram of a circuit consisting of a battery of three cells of $2\mathrm{V}$ each, a $5\mathrm{\Omega }$ resistor, an $8\mathrm{\Omega }$ resistor, and a $12\mathrm{\Omega }$ resistor, and a plug key, all connected in series.

Redraw the circuit of Question $1$, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the $12\mathrm{\Omega }$ resistor. What would be the readings in the ammeter and the voltmeter?

Answer 1

Step 1: Given data

It is given that,

${\mathrm{R}}_1=5\mathrm{\Omega }$,

${\mathrm{R}}_2=8\mathrm{\Omega }$,

${\mathrm{R}}_3=12\mathrm{\Omega }$

These are three resistors are connected in series.

The ammeter represents the amount of current in the circuit and the voltmeter represents the amount of voltage (potential difference) across $12\mathrm{\Omega }$.

We have to make a diagram of circuit and we need to find the current in the circuit $\mathrm{I}$ and potential difference across $12\mathrm{\Omega }$ resistor.

Step 2. Formula to be used

Ohm's law states that the voltage or potential difference between two points is directly proportional to the current or electricity which passing through the resistance, and directly proportional to the resistance of the circuit.

So, the formula of ohm's law is,

$\mathrm{V}=\mathrm{IR}$

Here, $\mathrm{V}$ is voltage, $\mathrm{I}$ is current and $\mathrm{R}$ is resistance.

Step 3: Draw a schematic diagram of a circuit

The diagram of the circuit is,

Step 4. Determine the resistance of the circuit.

From the given,

${\mathrm{R}}_1=5\mathrm{\Omega }$

${\mathrm{R}}_2=8\mathrm{\Omega }$

${\mathrm{R}}_3=12\mathrm{\Omega }$

Since the resistance are in series therefore the net resistance is $\mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3$

$=5+8+12$

$=25\mathrm{\Omega }$

Step 5. Find the current in the circuit

From the above diagram, because of the three cells of $2\mathrm{V}$ each, the potential difference across circuit is,

Potential difference across circuit $=2+2+2$

V $=6\mathrm{V}$

According to Ohm’s law,

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

Put the value of net voltage and equivalent resistance, we get,

$=\frac{6}{25}\mathrm{A}$

$=0.24\mathrm{A}$

Step 6. Find the potential difference across $12\mathrm{\Omega }$ resistor.

So, in series combination, current flowing through all resistors is same.

$\mathrm{V}=\mathrm{IR}$

Here,

$\mathrm{R}=12\mathrm{\Omega }$

$\mathrm{I}=0.24\mathrm{A}$

Put the above calculated value in the above formula, we get,

$=0.24\times 12$

$=2.88\mathrm{V}$

Hence, reading in the ammeter is $0.24\mathrm{A}$ and in voltmeter is $2.88\mathrm{V}$.