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Chapter 26: Problem 5

In these Problems neglect the internal resistance of a battery unless the Problem refers to it. (I) \(\mathrm{A} 650-\Omega\) and a \(2200-\Omega\) resistor are connected in series with a 12-V battery. What is the voltage across the \(2200-\Omega\) resistor?

Short Answer

Expert verified
The voltage across the 2200-惟 resistor is approximately 9.26 V.

Step by step solution

01

Identify Ohm's Law

Ohm's Law is crucial here, and it states \( V = I \times R \). This law will help us calculate the total current in the circuit and subsequently find the voltage across the \( 2200-\Omega \) resistor.
02

Calculate Total Resistance

Since the resistors are in series, the total resistance \( R_t \) is the sum of each resistor. \[ R_t = 650 \Omega + 2200 \Omega = 2850 \Omega \]
03

Calculate Total Current

Using Ohm's Law, we find the total current \( I \) in the circuit, where \( V_t \) is the voltage of the battery (12 V). \[ I = \frac{V_t}{R_t} = \frac{12 \text{ V}}{2850 \Omega} \approx 0.00421 \text{ A} \]
04

Calculate Voltage Across the 2200-惟 Resistor

Now, use Ohm's Law again for the \( 2200-\Omega \) resistor. Calculate the voltage \( V_{2200} \) across it: \[ V_{2200} = I \times R_{2200} = 0.00421 \text{ A} \times 2200 \Omega \approx 9.26 \text{ V} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

series circuit
A series circuit is one in which components are connected along a single path. This means the same current flows through all components of the circuit. When connecting resistors in series, like in our exercise, the total resistance is the sum of the individual resistances. In a series circuit:
  • The total resistance \( R_t \) is calculated by adding up each resistance: \( R_t = R_1 + R_2 + \ldots + R_n \).
  • The current \( I \) is constant through all components; it doesn't split like it might in a parallel circuit.
  • The voltage drop across each component can vary and depends on the resistance of each component.
Understanding this is crucial because it allows us to calculate the total resistance and find out how the voltage is distributed across each part of the circuit. For example, in our exercise, the resistors are in series, resulting in a single path for current to flow through both resistors before returning to the battery.
resistors
Resistors are components used to resist the flow of electric current in a circuit. They're key for controlling the voltage and current within electronics. When several resistors are placed in a circuit:
  • In a series circuit like in our example, resistors combine their resistances, affecting how current flows overall.
  • Each resistor will have a voltage drop across it, determined by Ohm's Law and the current flowing through it.
Ohm's Law, which states \( V = I \times R \), is used to calculate these effects in circuits. In our problem, we calculated the equivalent resistance of a \(650 \Omega\) and \(2200 \Omega\) resistor in series, then used it to determine the current. By understanding the role of resistors, we accurately determine how the voltage from a power source (like a battery) is distributed in a circuit.
voltage calculation
Voltage calculation is an essential part of circuit analysis, determining how much potential energy is being supplied to each component. Using Ohm's Law, we can find out how the total voltage is distributed among the components in a circuit.
  • In our exercise, the total voltage available is from a 12 V battery.
  • Using the total resistance, we calculate the current flowing through the circuit.
With the current known, we can calculate the voltage drop across any resistor by multiplying the current \( I \) by the resistor's resistance \( R \). Specifically, we determined that the voltage across the \(2200\,\Omega\) resistor is about \(9.26\,\text{V}\). This precise voltage calculation ensures we know how much voltage each part of the circuit receives, which is critical for understanding and designing electronic systems.

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Most popular questions from this chapter

(II) Three \(1.70-\mathrm{k} \Omega\) resistors can be connected together in four different ways, making combinations of series and/or parallel circuits. What are these four ways, and what is the net resistance in each case?

(II) A 1.5-V dry cell can be tested by connecting it to a lowresistance ammeter. It should be able to supply at least 25 A. What is the internal resistance of the cell in this case, assuming it is much greater than that of the ammeter?

(II) The performance of the starter circuit in an automobile can be significantly degraded by a small amount of corrosion on a battery terminal. Figure 38 depicts a properly functioning circuit with a battery \((12.5-\mathrm{V}\) emf, \(0.02-\Omega\) internal resistance \()\) attached via corrosion-free cables to a starter motor of resistance \(R_{\mathrm{S}}=0.15 \Omega\) Suppose that later, corrosion between a battery terminal and a starter cable introduces an extra series resistance of just \(R_{C}=0.10 \Omega\) into the circuit as suggested in Fig. 38 \(\mathrm{b}\) . Let \(P_{0}\) be the power delivered to the starter in the circuit free of corrosion, and let \(P\) be the power delivered to the circuit with corrosion. Determine the ratio \(P / P_{0}\) .

(II) A parallel-plate capacitor is filled with a dielectric of dielectric constant \(K\) and high resistivity \(\rho\) (it conducts very slightly). This capacitor can be modeled as a pure capacitance \(C\) in parallel with a resistance \(R\). Assume a battery places a charge \(+Q\) and \(-Q\) on the capacitor's opposing plates and is then disconnected. Show that the capacitor discharges with a time constant \(\tau=K \varepsilon_{0} \rho\) (known as the dielectric relaxation time). Evaluate \(\tau\) if the dielectric is glass with \(\rho=1.0 \times 10^{12} \Omega \cdot \mathrm{m}\) and \(K=5.0\).

(II) An \(R C\) series circuit contains a resistor \(R=15 \mathrm{k} \Omega\), a capacitor \(C=0.30 \mu \mathrm{F},\) and a battery of emf \(\mathscr{E}=9.0 \mathrm{~V}\) Starting at \(t=0\), when the battery is connected, determine the charge \(Q\) on the capacitor and the current \(I\) in the circuit from \(t=0\) to \(t=10.0 \mathrm{~ms}\) (at 0.1-ms intervals). Make graphs showing how the charge \(Q\) and the current \(I\) change with time within this time interval. From the graphs find the time at which the charge attains \(63 \%\) of its final value, \(C^{\mathscr{C}},\) and the current drops to \(37 \%\) of its initial value, \(\mathscr{E} / R\).
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