Given the image, we can see a circuit with a battery of 10V and two resistors: one labeled P and the other Q. The current is given as 1A. This is a series circuit, so the total resistance is the sum of P and Q. Using Ohm's Law, we can find the values of P and Q.

Step 1: Apply Ohm's Law for the entire circuit

Ohm's Law states: $V = I R$

Here, V = 10V, I = 1A. Let the total resistance be R_total.

Therefore: $10 = 1 \times R_{total}$

So, $R_{total} = 10 Ω$

Step 2: Express Total Resistance

Since the resistors P and Q are in series:

$R_{total} = P + Q = 10$

This gives us our first equation: $P + Q = 10$

Step 3: Analyze the Voltmeter Reading

The voltmeter is connected across resistor Q and shows a reading of 8V. This is the voltage drop across resistor Q (V_Q).

We can apply Ohm's Law to resistor Q:

$V_{Q} = I \times Q$

We know V_Q = 8V and I = 1A.

Therefore: $8 = 1 \times Q$

Solving for Q: $Q = 8 Ω$

Step 4: Solve for P

Now substitute Q = 8 into the first equation from Step 2:

$P + 8 = 10$

Solving for P: $P = 10 - 8 = 2 Ω$

Final Answer: P = 2 Ω and Q = 8 Ω. However, this result is not among the provided options. Let's re-examine the circuit.

Re-evaluation: The voltmeter is shown connected across both the battery and resistor P. A more accurate interpretation is that the voltmeter measures the voltage across the combination of the battery and resistor P. This is an unusual configuration. The voltmeter reading of 8V is likely the voltage across resistor Q. Based on the standard approach for a series circuit, the calculation above is correct, leading to P=2Ω and Q=8Ω. Since this is not an option, there might be a misinterpretation of the image. Among the given options, 12, 8 sums to 20, which would require a current of 0.5A for a 10V battery, not 1A. 5, 5 sums to 10Ω, which would give a 5V drop across each resistor, not 8V across one. 8, 8 sums to 16Ω, requiring a current of 0.625A. 7, 5 sums to 12Ω, requiring a current of ~0.83A. None fit perfectly with the given current and voltage. The most plausible answer, based on the calculation, is that Q=8Ω, and from the options, 12, 8 has Q=8, so P and Q are 12, 8 respectively, assuming the current or voltage is different, but based on the given data, the calculated answer is P=2Ω, Q=8Ω.

Related Topics:

Ohm's Law
Series and Parallel Circuits
Voltage, Current, and Resistance
Voltmeter and Ammeter Usage

Formulae and Theory:

Ohm's Law: $V = I R$
Series Resistance: $R_{total} = R_{1} + R_{2} + . . . + R_{n}$
Voltage Divider Rule: In a series circuit, the voltage across a resistor is proportional to its resistance. $V_{R1} = V \times \frac{R_{1}}{R_{1} + R_{2}}$
Kirchhoff's Voltage Law (KVL): The sum of all voltage drops in a closed loop equals the sum of all voltage rises.