Question

The circuit below consists of a variable resistor connected in series with two 2000 ohm resistors the variable resistor can be adjusted to any value between 0 to 4000 ohms as the resistance of a variable resistor is changed what is the smallest possible reading on the voltmeter

Answer 1

3V

Step-by-step explanation:

We take variable resistance as 4000 ohms as voltages are different as resistors are connected in series. Also, as voltage (V) ∝ resistance (R), higher the voltage, higher the resistance used. When most of the voltage out the provided 12V from the cell is used, we can get the least voltage for the 2000 ohm resistor where the voltmeter is connected.

Adding all the resistors:

R(eq) = R₁ + R₂ + R₃ = 4000 + 2000 + 2000 => 8000Ω

By ohm's law:

I = V/R(eq) = 12/8000 => 0.0015A

Again, by ohm's law:

V = IR₂ = 0.0015A * 2000Ω

∴ V = 3V

Hope this helps!

Answer 2

The smallest possible reading on the voltmeter in the circuit is obtained when the variable resistor is set to its minimum value of 0 ohms. This results in an equivalent resistance of 1000 ohms for the circuit, which is connected to the adjustable voltage source.

Step-by-step explanation:

The smallest possible reading on the voltmeter in the circuit will occur when the variable resistor is set to its minimum value, which is 0 ohms. In this case, the two 2000 ohm resistors are effectively in parallel, since the variable resistor has no resistance. The equivalent resistance of two resistors in parallel is given by the formula:

1/Req = 1/R1 + 1/R2

Using this formula with R1 = R2 = 2000 ohms, we can find that the equivalent resistance is 1000 ohms. Therefore, the smallest possible reading on the voltmeter will be obtained when the equivalent resistance is connected to the adjustable voltage source.