Study the following electric circuit. Find the reading of

(a) the ammeter and

(b) the voltmeter

(a) In the given circuit, resistances are connected in series, hence, equivalent resistances (R_e) = R₁ + R₂ = 4 + 2 = 6 Ω

Potential difference (V) = 3 V

Current (I) = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

$3 = I \times 6 \\[0.5em] \Rightarrow I = \dfrac{3}{6} \\[0.5em] \Rightarrow I = 0.5 A$

As in series combination, current flowing through each component is same and is equal to the total current flowing in the circuit hence, the reading of ammeter = 0.5 A

(b) Reading of voltmeter = Potential difference across 2 Ω resistor.

Current flowing through bulb = 0.5 A

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 0.5 x 2 = 1 V

Hence, the reading of voltmeter = 1 V