Question

What would be the reading of ammeter and voltmeter in the given circuit?

Answer 1

Hint: To solve this question we need to find out the total resistance in the circuit. Then using Ohm's law we can determine the current in the circuit, which will be the reading of the ammeter. The reading of the voltmeter can be determined by finding the potential difference across the $2\mathrm{\Omega }$ resistance by using the value of current in the circuit.

Formula used: The formula which is used in this question is given by
 $V=IR$, here $V$ is the potential difference across a resistance $R$ through which a current $I$ is flowing.

Complete step by step answer
In the given circuit, the two resistances are connected in series combination. So the net resistance of the circuit becomes
 $R=1\mathrm{\Omega }+2\mathrm{\Omega }$
 $\Rightarrow R=3\mathrm{\Omega }$
The battery is of emf $3\text{V}$. So the current flowing in the circuit is given by the Ohm’s law as
 $3=3I$
 $\Rightarrow I=1\text{A}$ …………………...(1)
The ammeter is connected in series with the circuit. So its reading will show the value of the current flowing in the circuit. Hence, the reading of the ammeter is equal to $1\text{A}$.
Now, the voltmeter is connected in parallel combination with the $2\mathrm{\Omega }$ resistance. So its reading will represent the potential difference across the $2\mathrm{\Omega }$ resistance. The potential difference across the $2\mathrm{\Omega }$ resistance is given by the Ohm’s law as
 $V=IR$
Substituting (1) and $R=2\mathrm{\Omega }$ above, we get
 $V=1\times 2$
 $\Rightarrow V=2\text{V}$
Hence, the voltmeter will show a reading of $2\text{V}$.

Note
In these types of questions, we should not forget to notice the switch given in the circuit diagram. The switch in the given question is closed, so we ignored it. But it may happen that the switch is open and you ignore it and proceed with the calculations. If the switch is open, then the reading of both the ammeter and the voltmeter will become zero.