Question

A cell, an ammeter, and a voltmeter are all connected in series. The ammeter reads a current $I$ and the voltmeter a potential difference $V$. If a torch bulb is connected across the voltmeter, then.
(A) Both $I$ and $V$ will increase
(B) Both $I$ and $V$ will decrease
(C) $I$ will increase but $V$ will decrease
(D) $I$ will decrease but $V$ will increase

Answer 1

Hint
The current remains constant in series combination. Whereas potential difference remains constant in parallel combination. And, when resistances are connected in parallel combination, the resultant of the setup decreases.

Complete step by step solution
The effective resistance for parallel combination is given by,
${{\rm{R}}_{{\rm{effective}}}} = \dfrac{{{{\rm{R}}_1}{{\rm{R}}_2}}}{{{{\rm{R}}_1} + {{\rm{R}}_2}}}$
The resistance of the torch bulb will be in parallel combination with the resistance of the voltmeter which will result in low effective resistance. Now, according to ohm’s law, the relation between the resistance and the current is given by,
${\rm{I}} = \dfrac{{\rm{V}}}{{\rm{R}}}$
Since the resistance is decreased; therefore, the current in the circuit will increase. Thus, the reading in the ammeter will increase i.e. the current across the ammeter will increase. So, the potential difference across the voltmeter will decrease.

Therefore, (C) I will increase but V will decrease; is the required answer statement.

Additional information
According to the ohm’s law, at the given temperature, the electric potential difference across the ends of the circuit is directly proportional to the current flowing in the circuit.
i.e. ${\rm{V}} \propto {\rm{I}}$
${\rm{V}} = {\rm{RI}}$.

Note
The voltmeter is always connected in parallel across the points in the circuit where the potential difference is to be calculated. Whereas ammeters can be connected in series where current is to be calculated.