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Chapter 25: Problem 75

Three resistors are connected to a power supply with \(V=110 . \mathrm{V}\) as shown in the figure a) Find the potential drop across \(R_{3}\) b) Find the current in \(R_{1}\). c) Find the rate at which thermal energy is dissipated from \(R_{2}\).

Short Answer

Expert verified
The formula to calculate the equivalent resistance of resistors connected in parallel is: $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$$

Step by step solution

01

Analyze the circuit

First, let's identify the type of connections for the resistors. \(R_1\) and \(R_3\) are connected in series, and this series connection is in parallel with \(R_2\).
02

Calculate the equivalent resistance

To find the potential drop across \(R_3\) and the current in \(R_1\), we need to find the equivalent resistance of the parallel connection, which we will call \(R_{eq}\). In order to find \(R_{eq}\), we first need to find the equivalent resistance of the series connection of \(R_1\) and \(R_3\). The equivalent resistance of resistors in series is given by: $$R_{eq1} = R_1 + R_3$$ Now we can find the equivalent resistance of \(R_{eq1}\) and \(R_2\) in parallel using the formula: $$\frac{1}{R_{eq}} = \frac{1}{R_{eq1}} + \frac{1}{R_2}$$
03

Find the total current coming from the power supply

Now that we have found \(R_{eq}\), we can use Ohm's law to find the total current coming from the power supply. Ohm's law states that: $$I = \frac{V}{R}$$ Here, the total current \(I\) is equal to the voltage \(V\) divided by the equivalent resistance \(R_{eq}\). Substitute the values and calculate the current.
04

Find the potential drop across \(R_3\)

To find the potential drop across \(R_3\), first, we need to find the current flowing through \(R_1\) and \(R_3\). Since both are in series, the current flowing through them is the same as the total current coming from the power supply. After finding the current flowing through \(R_3\), use the Ohm's law formula to find the potential drop across it: $$V_{R_3} = I_{R_3} \times R_3$$
05

Find the current flowing through \(R_1\)

As mentioned earlier, the current flowing through \(R_1\) is the same as the total current coming from the power supply. Therefore, the current in \(R_1\) is equal to the current calculated in step 3.
06

Calculate the rate of thermal energy dissipation

To find the rate at which thermal energy is dissipated from \(R_2\), we first need to find the current flowing through \(R_2\). Since \(R_2\) is in parallel with the series connection of \(R_1\) and \(R_3\), the current flowing through it can be found using the formula: $$I_{R_2} = \frac{V}{R_2}$$ Next, use the formula for power to find the rate of thermal energy dissipation: $$P_{R_2} = I_{R_2}^2 \times R_2$$

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Most popular questions from this chapter

If the current through a resistor is increased by a factor of \(2,\) how does this affect the power that is dissipated? a) It decreases by a factor of 4 . b) It increases by a factor of 2 . c) It decreases by a factor of 8 . d) It increases by a factor of 4 .

An infinite number of resistors are connected in parallel. If \(R_{1}=10 \Omega, R_{2}=10^{2} \Omega, R_{3}=10^{3} \Omega,\) and so on, show that \(R_{e q}=9 \Omega\).

A copper wire has a diameter \(d_{\mathrm{Cu}}=0.0500 \mathrm{~cm}\) is \(3.00 \mathrm{~m}\) long, and has a density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). As shown in the figure, the copper wire is attached to an equal length of aluminum wire with a diameter \(d_{\mathrm{A} \mathrm{I}}=0.0100 \mathrm{~cm}\) and density of charge carriers of \(6.02 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). A current of 0.400 A flows through the copper wire. a) What is the ratio of the current densities in the two wires, \(J_{\mathrm{Cu}} / J_{\mathrm{Al}} ?\) b) What is the ratio of the drift velocities in the two wires, \(v_{\mathrm{d}-\mathrm{Cu}} / v_{\mathrm{d}-\mathrm{Al}} ?\)

A water heater consisting of a metal coil that is connected across the terminals of a 15 -V power supply is able to heat \(250 \mathrm{~mL}\) of water from room temperature to boiling point in \(45 \mathrm{~s}\). What is the resistance of the coil?

Two resistors with resistances \(R_{1}\) and \(R_{2}\) are connected in parallel. Demonstrate that, no matter what the actual values of \(R_{1}\) and \(R_{2}\) are, the equivalent resistance is always less than the smaller of the two resistances.
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