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A \(150-\mathrm{V}\) voltmeter has a resistance of \(30,000\) . \(\Omega\) . When connected in series with a large resistance \(R\) across a \(110-\mathrm{V}\) line, the meter reads 68 \(\mathrm{V}\) . Find the resistance \(R\) .

Short Answer

Expert verified
The resistance \(R\) is approximately \(18,529.41 \, \Omega\).

Step by step solution

01

Understand the Circuit Setup

The circuit involves a voltmeter with a resistance of \(30,000\, \Omega\), measuring a voltage drop across a part of the circuit, while connected in series with a resistance \(R\). The total voltage supplied is \(110\, \mathrm{V}\), and the voltmeter reads \(68\, \mathrm{V}\).
02

Calculate Voltage Across Resistance R

The total voltage across the series circuit is \(110\, \text{V}\). Since the voltmeter reads \(68\, \text{V}\), this means the remaining voltage across resistance \(R\) is \(110\, \text{V} - 68\, \text{V} = 42\, \text{V}\).
03

Apply Ohm's Law to Find Total Current

Using Ohm's Law, the current \(I\) through the circuit can be expressed as \(I = \frac{V}{R}\). For the voltmeter, \(I = \frac{68}{30,000}\, \text{A}\).
04

Use Current to Find Resistance R

The current through the voltmeter is the same as the current through \(R\) because they are in series. Therefore, we can use \(I = \frac{42}{R}\) and set it equal to the previous current calculation to find \(R\).
05

Solve for R

By equating the currents: \(\frac{68}{30,000} = \frac{42}{R}\). Cross-multiplying gives: \(68R = 30,000 \times 42\). Solving for \(R\) yields \(R = \frac{30,000 \times 42}{68}\). Simplifying gives \(R \approx 18,529.41\, \Omega\).

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Key Concepts

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

Voltmeter Resistance
A voltmeter is an instrument used to measure the electrical potential difference between two points in an electric circuit. One critical aspect of a voltmeter is its internal resistance. For accurate measurements, a voltmeter should have a very high resistance, ideally infinite. This high resistance ensures minimal current flows through the voltmeter, preventing it from significantly altering the circuit it is measuring. In our scenario, the voltmeter has a resistance of 30,000 ohms. This resistance is substantial enough to ensure that the current passing through it is small. Limiting the current is essential because any notable change can affect the accuracy of the voltage measurement, leading to incorrect readings. Remember: - The higher the voltmeter resistance, the better it is for precise voltage readings. - A voltmeter's high resistance minimizes its impact on the circuit.
Series Circuit
A series circuit is a type of electrical circuit in which components are connected end-to-end, so the same current flows through each component without branching. The total resistance ( R_{total} ) in a series circuit is simply the sum of the individual resistances:\[ R_{total} = R_1 + R_2 + ... + R_n \]In our case, the components in series are the voltmeter and an unknown resistor, R. The total resistance of this series circuit can be calculated once both resistances are known.Benefits of a series circuit include:
  • Simplicity in design and construction.
  • Ensures uniform current distribution.
However, if one component fails, the entire circuit is disrupted, as the current has only one path to follow.
Voltage Drop
Voltage drop refers to the reduction in voltage as it moves through components of a circuit. In a series circuit, the voltage drop across each component depends on its resistance relative to the total resistance.Using Ohm's Law, the voltage drop ( V_{drop} ) across a resistor within a circuit is given by:\[ V_{drop} = I imes R \]Where I is the circuit's current, and R is the resistance of the component. In the problem, since the total circuit voltage is 110 V and the voltmeter reads 68 V, it indicates that the voltage drop across the resistor R is 42 V:\[ 110 ext{ V} - 68 ext{ V} = 42 ext{ V} \]Understanding voltage drops is crucial because they help us analyze circuits and find how voltage is divided across the components.
Electrical Resistance Calculation
Electrical resistance is a measure of the opposition to the flow of current in an electrical circuit. Knowing how to calculate it is essential for analyzing circuits accurately. In a series circuit, all components share the same current. Using Ohm's Law, which states that V = I imes R , we can calculate unknowns in the circuit. For the given exercise, the current through the voltmeter and resistance R is the same, enabling us to equate two expressions of I :\[ I = \frac{68}{30,000} = \frac{42}{R} \]Solving the equation allows us to find R:\[ R = \frac{30,000 \times 42}{68} \]This formula lets us determine that the unknown resistance R is approximately 18,529.41 ohms.Key points for calculating resistance:
  • Use Ohm's Law to understand relationships between voltage, current, and resistance.
  • Ensure units are consistent when performing calculations.
  • Consider the configuration of the circuit when solving for unknown elements.

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Most popular questions from this chapter

A circuit consists of a series combination of \(6.00-\mathrm{k} \Omega\) and 5.00- \(\mathrm{k} \Omega\) resistors connected across a \(50.0-\mathrm{V}\) battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the \(5.00-\mathrm{k} \Omega\) resistor using a voltmeter having an internal resistance of 10.0 \(\mathrm{k} \Omega\) (a) What potential difference does the voltmeter measure across the \(5.00-\mathrm{k} \Omega\) resistor? (b) What is the true potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?

A \(224-\Omega\) resistor and a \(589-\Omega\) resistor are connected in series across a \(90.0-\mathrm{V}\) line. (a) What is the voltage across each resistor? (b) A voltmeter connected across the \(224-\Omega\) resistor reads 23.8 \(\mathrm{V}\) . Find the voltmeter resistance. (c) Find the reading of the same volt- meter if it is connected across the \(589-\Omega\) resistor. (d) The readings on this voltmeter are lower than the "true" voltages (that is, without the voltmeter present). Would it be possible to design a voltmeter that gave readings higher than the "true" voltagcs? Explain.

A capacitor is charged to a potential of 12.0 \(\mathrm{V}\) and is then connected to a voltmeter having an internal resistance of 3.40 \(\mathrm{M} \Omega\) . After a time of 4.00 s the voltmeter reads 3.0 \(\mathrm{V}\) . What are (a) the capacitance and \((\mathrm{b})\) the time constant of the circuit?

A certain galvanometer has a resistance of 65.0\(\Omega\) and deflects full scale with a current of 1.50 \(\mathrm{mA}\) in its coil. This is to be replaced with a second galvanometer that has a resistance of 38.0\(\Omega\) and deflects full scale with a current of 3.60\(\mu A\) in its coil. Devise a circuit incorporating the second galvanometer such that the equivalent resistance of the circuit equals the resistance of the first galvanometer, and the second galvanometer deflects full scale when the current through the circuit equals the full-scale current of the first galvanometer.

The heating element of an electric stove consists of a heater wire embedded within an electrically insulating material, which in turn is inside a metal casing. The heater wire has a resistance of 20\(\Omega\) at room temperature \(\left(23.0^{\circ} \mathrm{C}\right)\) and a temperature coefficient of resistivity \(\alpha=2.8 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\) . The heating element oper- ates from a 120 \(\mathrm{V}\) line. (a) When the heating element is first turned on, what current does it draw and what electrical power does it dis- sipate? (b) When the heating element has reached an operating temperature of \(280^{\circ} \mathrm{C}\left(536^{\circ} \mathrm{F}\right),\) what current does it draw and what electrical power does it dissipate?

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