Question:

A wire of resistance 5.0 ohms is connected to a battery with an emf of 12.0 V. The wire is then connected in parallel with another wire of resistance R. When the circuit has reached steady state, an ammeter in the circuit reads 2.0 A. Determine the resistance R.

(a) 12.0 ohms

(b) 4.0 ohms

(c) 3.0 ohms

(d) 6.0 ohms

Answer and Explanation:

We can solve this problem by using the concept of equivalent resistance for resistors connected in parallel. The equivalent resistance for resistors connected in parallel is given by the formula:

1/Req = 1/R1 + 1/R2 + 1/R3 + ...

In this case, the wire with resistance 5.0 ohms is connected in parallel with another wire of resistance R. Let's assume the equivalent resistance of the circuit is Req.

1/Req = 1/5.0 + 1/R

To find Req, we can rearrange the formula:

Req = R * (5.0 / (R + 5.0))

Now, we can use Ohm's Law to find the current in the circuit. Ohm's Law states that:

V = I * R

where V is the voltage (emf of the battery), I is the current, and R is the resistance. Since the wires are connected in parallel, the voltage across both wires is the same and equal to the emf of the battery.

12.0 V = 2.0 A * (5.0 ohms + R)

Let's solve this equation for R:

12.0 V = 10.0 A * (5.0 ohms + R)

2.0 A * (5.0 ohms + R) = 12.0 V

5.0 ohms + R = 12.0 V / 2.0 A

5.0 ohms + R = 6.0 ohms

R = 6.0 ohms - 5.0 ohms

R = 1.0 ohms

Therefore, the resistance R is 1.0 ohms.

Since none of the given options match the calculated value, the closest option is (d) 6.0 ohms.