Question

A circuit consists of two \(1.00-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-V\) battery. a) Calculate the current flowing through each resistor. b) A student trying to measure the current flowing through one of the resistors inadvertently connects an ammeter in parallel with that resistor rather than in series with it. How much current will flow through the ammeter, assuming that it has an internal resistance of \(1.00 \Omega ?\)

Short answer

Answer: a) The current flowing through each resistor in the series circuit is 6.00 mA. b) In the modified circuit, the current flowing through the ammeter, connected in parallel with one of the resistors, is approximately 11.9 mA.

Step-by-step solution

Find the total resistance

In a series circuit, the total resistance is simply the sum of the individual resistances. There are two \(1.00 \text{k}\Omega\) resistors in series, so the total resistance can be found as: \(R_{total} = R_1 + R_2 = 1.00\text{k}\Omega + 1.00\text{k}\Omega = 2.00\text{k}\Omega\)

Calculate the current flowing in the circuit

We can now use Ohm's Law to find the current flowing in the circuit: \(I = \frac{V}{R}\) Where \(I\) is the current, \(V\) is the voltage, and \(R\) is the resistance of the circuit. Here, \(V = 12.0 V\) and \(R =2.00\text{k}\Omega\). So: \(I = \frac{12.0V}{2.00\text{k}\Omega} = 0.006\text{A}\) or \(6.00\text{mA}\)

Answer to part (a)

Since the resistors are connected in series, the current flowing through each resistor is the same. Therefore, the current flowing through each resistor is also \(0.006\text{A}\) or \(6.00\text{mA}\).

Redraw the circuit with the ammeter connected in parallel

Now, let's consider the case where the ammeter is connected in parallel with one of the resistors rather than in series. First, we need to redraw the circuit with this setup. The ammeter has an internal resistance of \(1.00\Omega\). Let resistor \(R_1\) remain connected in the same way but let resistor \(R_2\) be parallel with the ammeter ( \(R_A = 1.00 \Omega\)). The new circuit will consist of resistor \(R_1\) in series with the parallel combination of \(R_2\) and \(R_A\).

Calculate the equivalent resistance of \(R_2\) and \(R_A\) in parallel

To find the equivalent resistance for resistors connected in parallel, we use the formula: \(\frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_A}\) Where \(R_{eq}\) is the equivalent resistance, \(R_2\) is the resistance of the second resistor, and \(R_A\) is the internal resistance of the ammeter. Plug in the values: \(\frac{1}{R_{eq}} = \frac{1}{1.00\text{k}\Omega} + \frac{1}{1.00\Omega}\) First, convert the \(1.00\text{k}\Omega\) into \(\Omega\): \(\frac{1}{R_{eq}} = \frac{1}{1000\Omega} + \frac{1}{1.00\Omega}\) Now, solve for \(R_{eq}\): \(R_{eq} = \frac{1}{\frac{1}{1000\Omega} + \frac{1}{1.00\Omega}} \approx 0.999\Omega\)

Calculate the total resistance in the new circuit

Now, we need to find the total resistance in the new circuit. The total resistance is the sum of \(R_1\) and \(R_{eq}\): \(R_{total} = R_1 + R_{eq} = 1.00\text{k}\Omega + 0.999\Omega \approx 1.00\text{k}\Omega\)

Calculate the current in the new circuit

Using Ohm's Law, we can calculate the current in the new circuit: \(I_{total} = \frac{V}{R_{total}} = \frac{12.0V}{1.00\text{k}\Omega} \approx 0.012\text{A}\) or \(12.0\text{mA}\)

Calculate the current flowing through the ammeter

The current is now divided between \(R_2\) and the ammeter (\(R_A\), connected in parallel). To calculate the current flowing through the ammeter, we can use the current divider rule: \(I_{A} = I_{total} \frac{R_2}{R_2 + R_A} = 0.012\text{A} \frac{1.00\text{k}\Omega}{1.00\text{k}\Omega + 1.00\Omega} \approx 0.011\text{A}\) or \(11.9\text{mA}\)

Answer to part (b)

The current flowing through the ammeter, when it is connected in parallel with one of the resistors, is \(0.011\text{A}\) or \(11.9\text{mA}\).