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Chapter 18: Problem 42

If a \(93.5-\mathrm{V}\) emf is connected to the terminals \(A\) and \(B\) and the current in the \(4.0-\Omega\) resistor is \(17 \mathrm{A},\) what is the value of the unknown resistor \(R ?\)

Short Answer

Expert verified
Question: Calculate the value of the unknown resistor R in a circuit with a 93.5 V emf source, a 4.0 惟 resistor with 17 A current passing through it, and an unknown resistor in parallel with the given resistor. Answer: The value of the unknown resistor R is approximately 14.67 惟.

Step by step solution

01

Identify the problem components

We have the following components in the problem: 1. Emf source (E) of 93.5 V 2. Resistor 1 (R1) of 4.0 惟 3. Resistor 2 (R2) with an unknown value (R) 4. Current (I) in R1 is 17 A
02

Determine the voltage across the resistors

Since the emf source is connected across the resistors, the voltage across each resistor will be the same as the emf voltage. Therefore, the voltage across R1 (V1) and R2 (V2) is 93.5 V.
03

Find the current through the 4.0 惟 resistor

We are given that the current through R1 is 17 A.
04

Determine the current through the unknown resistor

The current through the unknown resistor R2 (I2) is unknown. However, since R1 and R2 are connected in parallel, the voltage across both resistors will be the same.
05

Use Ohm's Law to find the value of the unknown resistor

Ohm's Law states that \(V = IR\). We know the voltage across R2 (V2) and the current through R1 (I1). We can rearrange the formula to solve for the unknown resistor R: \(R2 = \frac{V2}{I2}\) We know that V2 is 93.5 V. We need to find I2.
06

Apply Kirchhoff's Current Law

Kirchhoff's Current Law states that the sum of currents entering a node must equal the sum of currents leaving that node. In our case, this means that the current entering the circuit (I0) divided between the resistors (I1 and I2) must sum up to I0: \(I0 = I1 + I2\) We know that I1 is 17 A. We can rearrange the formula to solve for I2: \(I2= I0 - I1\)
07

Find the total current in the circuit

To find the total current I0, we can use Ohm's law for the entire circuit: \(I0 = \frac{E}{R_{eq}}\) where Re is the equivalent resistance of the circuit. Since R1 and R2 are in parallel, we can use the formula for resistors in parallel to find the equivalent resistance: \(\frac{1}{R_{eq}} = \frac{1}{R1}+\frac{1}{R2}\)
08

Calculate the equivalent resistance

Knowing R1, we can solve for Re: \(\frac{1}{R_{eq}} = \frac{1}{4.0\,\Omega}+\frac{1}{R}\) To solve this equation, we need to find the value of the total current (I0) first.
09

Find the total current (I0) using given values

We know the voltage across R1 (V1) and the current through R1 (I1). Using Ohm's Law: \(I0 = \frac{E}{R1} = \frac{93.5\,V}{4\,\Omega} = 23.375\,A\) Now we have the I0 value, we can find I2.
10

Calculate I2

Using the formula from Step 6: \(I2 = 23.375\,A - 17\,A = 6.375\,A\)
11

Find the value of the unknown resistor (R2)

Now we can plug the value of I2 back into the Ohm's Law formula from Step 5: \(R2 = \frac{V2}{I2} = \frac{93.5\,V}{6.375\,A} = 14.67\,\Omega\) Thus, the value of the unknown resistor R is approximately 14.67 惟.

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Most popular questions from this chapter

Near Earth's surface the air contains both negative and positive ions, due to radioactivity in the soil and cosmic rays from space. As a simplified model, assume there are 600.0 singly charged positive ions per \(\mathrm{cm}^{3}\) and 500.0 singly charged negative ions per \(\mathrm{cm}^{3} ;\) ignore the presence of multiply charged ions. The electric field is $100.0 \mathrm{V} / \mathrm{m},$ directed downward. (a) In which direction do the positive ions move? The negative ions? (b) What is the direction of the current due to these ions? (c) The measured resistivity of the air in the region is $4.0 \times 10^{13} \Omega \cdot \mathrm{m} .$ Calculate the drift speed of the ions, assuming it to be the same for positive and negative ions. [Hint: Consider a vertical tube of air of length \(L\) and cross-sectional area \(A\) How is the potential difference across the tube related to the electric field strength?] (d) If these conditions existed over the entire surface of the Earth, what is the total current due to the movement of ions in the air?
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