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Chapter 26: Problem 39

A circuit consists of two \(1.00-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) 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 \(1.0 \Omega ?\)

Short Answer

Expert verified
Answer: The current flowing through each resistor in the series circuit is \(6.00\,\mathrm{mA}\). When the ammeter is connected in parallel to one of the resistors, the current flowing through the ammeter is approximately \(12.0\,\mathrm{mA}\).

Step by step solution

01

Calculate the equivalent resistance of the circuit

To calculate the current flowing through each resistor, we first need to find the equivalent resistance of the circuit. Since the two resistors are in series, we can find the equivalent resistance by simply adding their values: \(R_{eq} = R_1 + R_2 = 1.00\,\mathrm{k\Omega} + 1.00\,\mathrm{k\Omega} = 2.00\,\mathrm{k\Omega}\)
02

Calculate current using Ohm's Law

Now that we have the equivalent resistance, we can use Ohm's Law to find the current flowing through the circuit. Ohm's Law states that \(V = IR,\) where V is the voltage, I is the current, and R is the resistance. In our case, we have: \(I = \dfrac{V}{R_{eq}} = \dfrac{12.0\,\mathrm{V}}{2.00\,\mathrm{k\Omega}} = 6.00\,\mathrm{mA}\) Since the resistors are in series, the current flowing through each resistor is the same, which is \(6.00\,\mathrm{mA}.\)
03

Calculate the equivalent resistance of the circuit with the ammeter

When the student connects the ammeter in parallel with one of the resistors, the equivalent resistance of the parallel combination of the resistor and the ammeter is: \(R_{par} = \dfrac{R_3 \cdot R_A}{R_3 + R_A} = \dfrac{1.00\,\mathrm{k\Omega} \cdot 1.0\,\mathrm{\Omega}}{1.00\,\mathrm{k\Omega} + 1.0\,\mathrm{\Omega}} \approx 1.0\,\mathrm{\Omega}\)
04

Calculate the new equivalent resistance of the circuit

Now, we have the old resistor replaced by the equivalent parallel resistance (including the ammeter). So, we can calculate the new equivalent resistance of the circuit by adding the second resistor and the equivalent parallel resistance: \(R_{eq2} = R_2 + R_{par} = 1.00\,\mathrm{k\Omega} + 1.0\,\mathrm{\Omega} \approx 1.00\,\mathrm{k\Omega}\)
05

Calculate the current using Ohm's Law with the new equivalent resistance

Now that we have the new equivalent resistance, we can use Ohm's Law again to find the current flowing through the circuit: \(I_2 = \dfrac{V}{R_{eq2}} = \dfrac{12.0\,\mathrm{V}}{1.00\,\mathrm{k\Omega}} = 12.0\,\mathrm{mA}\)
06

Calculate the current flowing through the ammeter

Since the ammeter is connected in parallel with one of the resistors, we need to find the current flowing through the ammeter using the current divider rule: \(I_A = I_2 \dfrac{R_3}{R_3 + R_A} = 12.0\,\mathrm{mA} \times \dfrac{1.00\,\mathrm{k\Omega}}{1.00\,\mathrm{k\Omega} + 1.0\,\mathrm{\Omega}} \approx 12.0\,\mathrm{mA}\) So, the current flowing through the ammeter is approximately \(12.0\,\mathrm{mA}\).

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Most popular questions from this chapter

An RC circuit has a time constant of 3.1 s. At \(t=0\), the process of charging the capacitor begins. At what time will the energy stored in the capacitor reach half of its maximum value?

Two resistors, \(R_{1}\) and \(R_{2},\) are connected in series across a potential difference, \(\Delta V_{0}\). Express the potential drop across each resistor individually, in terms of these quantities. What is the significance of this arrangement?

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