Consider the following series circuit consisting of a 12 V battery, two identical resistors (\(R_1\) and \(R_2\)), and an ammeter with a reading of 2 A. Voltmeters \(V_t\), \(V_1\), and \(V_2\) are connected over the battery and resistors.

Calculate:
1. The resistance of each resistor if the total resistance is 6 Ω;
2. The reading on \(V_1\) if the emf is 12 V;
3. The reading on \(V_2\) if the emf is 12 V.

Finding the Resistance of Each Resistor:
- Ohm's Law states that [tex]V = I \times R[/tex] where [tex]V[/tex] is the voltage, [tex]I[/tex] is the current, and [tex]R[/tex] is the resistance.
- Given that the total resistance [tex]R_{total}[/tex] is 6 Ω and the resistors are identical, they divide this total resistance equally.
- Therefore, the resistance of each resistor [tex]R_1[/tex] and [tex]R_2[/tex] will be half of the total resistance:
  [tex]R_1 = R_2 = \frac{R_{total}}{2} = \frac{6 \, \Omega}{2} = 3 \, \Omega[/tex]

The Reading on Voltmeter [tex]V_1[/tex]:
- Voltmeter [tex]V_1[/tex] measures the voltage across resistor [tex]R_1[/tex].
- Again applying Ohm's Law: [tex]V_1 = I \times R_1[/tex]
- Given that [tex]I = 2 \, A[/tex], and [tex]R_1 = 3 \, \Omega[/tex]:
  [tex]V_1 = 2 \, A \times 3 \, \Omega = 6 \, V[/tex]

The Reading on Voltmeter [tex]V_2[/tex]:
- Voltmeter [tex]V_2[/tex] measures the voltage across resistor [tex]R_2[/tex].
- We apply the same calculation as for [tex]V_1[/tex] since both resistors are identical and have the same current running through them:
  [tex]V_2 = I \times R_2 = 2 \, A \times 3 \, \Omega = 6 \, V[/tex]

In summary, the resistance of each resistor is 3 Ω, the reading on voltmeter [tex]V_1[/tex] is 6 V, and the reading on voltmeter [tex]V_2[/tex] is also 6 V, making the total voltage across the series circuit (and hence measured by [tex]V_t[/tex]) equal to the emf [tex]12 \, V[/tex] provided by the battery (since [tex]V_1 + V_2 = 12 \, V[/tex]). This verifies that our calculations are consistent, and the circuit is correctly analyzed.