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Chapter 30: Problem 27

. In Fig. \(30.11 .\) suppose that \(\varepsilon=60.0 \mathrm{~V}, R=240 \Omega .\) and \(L=0.160 \mathrm{H}\). With switch \(S_{2}\) open, switch \(S_{1}\) is left closed until a constant currcnt is cstablishcd. Then \(S_{2}\) is closcd and \(S_{1}\) opcned, taking the battery out of the circuit. (a) What is the initial current in the resistor, just after \(S_{2}\) is closed and \(S_{1}\) is opened? (b) What is the current in the resistor at \(t=4.00 \times 10^{-4} \mathrm{~s} ?\) (c) What is the potcntial differcnoe betwecn points \(b\) and \(c\) at \(t=4.00 \times 10^{-4} \mathrm{~s}\) ? Which point is at a higher potcntial? (d) How long does it take the current to decrease to half its initial value?

Short Answer

Expert verified
(a) The initial current is 0.25 A. (b) The current at \(t = 4.00 \times 10^{-4}~s\) is 0.182 A. (c) The potential difference \(V_{bc}\) is 43.7 V. Point \(b\) is at a higher potential. (d) It takes \(1.10 \times 10^{-3} ~s\) for the current to decrease to half its initial value.

Step by step solution

01

Calculate Initial Current

After the system reaches a steady state with the switch \(S_{1}\) closed, and before \(S_{2}\) is closed, the circuit simplifies to a simple Ohm's law problem. The current \(I_{0}\), given by Ohm's law, is \(I_{0} = \frac{蔚}{R} = \frac{60.0~V}{240~\Omega}\)
02

Calculate Current at Given Time

Once the \(S_{1}\) is opened and \(S_{2}\) is closed, the system becomes an RL-circuit. The current decay in such circuit is given by the formula \(I(t) = I_{0}e^{-\frac{t}{\tau}}\) , where \(\tau = \frac{L}{R}\) is the time constant of RL-circuit. Thus, first calculate \(\tau = \frac{L}{R} = \frac{0.160 H}{240 \Omega}\). Then, substitute \(I_{0}\), \(\tau\) and \(t = 4.00 \times 10^{-4}~s\) in the formula to find the current at the given time.
03

Calculate Potential Difference

The potential difference between points \(b\) and \(c\) can be calculated using the relationship \(V_{bc} = I(t)R\). Use the current calculated in Step 2 and the given resistance to calculate the potential difference. The polarity of the difference will indicate which point has higher potential.
04

Calculate time for Current to decrease by half

The formula for current decay in an RL-circuit can be used to find the time it takes for the current to decrease to half its initial value. Thus, by setting \(I(t) = \frac{I_{0}}{2}\), and then substituting \(I_{0}\) and \(\tau\), solve the equation to find the value of \(t\).

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Key Concepts

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

Ohm's Law
Understanding Ohm's Law is critical when it comes to analyzing electrical circuits. It is a fundamental principle in physics that relates the voltage (potential difference), current, and resistance. Defined by the formula
\( I = \frac{V}{R} \)
, Ohm's Law states that the current
\( I \)
flowing through a conductor between two points is directly proportional to the voltage
\( V \)
across the two points and inversely proportional to the resistance
\( R \)
of the conductor. In our exercise, we applied Ohm's Law to determine the initial current in the circuit with the formula
\( I_0 = \frac{\varepsilon}{R} \)
, where
\( \varepsilon \)
is the electromotive force (EMF) of the battery, and
\( R \)
is the resistance of the resistor. This principle lays the groundwork for analyzing more complex behaviors in electrical circuits, such as the decay of current in an RL-circuit.
Time Constant
The time constant, denoted as
\( \tau \)
, plays a pivotal role in the behavior of RL-circuits. It dictates the rate at which the current decays after the switch
\( S_1 \)
is opened. The time constant is the product of the inductance
\( L \)
and the resistance
\( R \)
, represented by the formula
\( \tau = \frac{L}{R} \)
. Simply put, it signifies the time it takes for the current to reduce to about 37% of its initial value. In our example, once we have calculated the time constant, we can predict how quickly the current will fall over time. The faster the decay, the smaller the time constant, demonstrating a strong linkage between the physical components of the circuit and their temporal behavior on current flow.
Potential Difference
In an electrical circuit, the term potential difference refers to the work done to move a charge from one point to another. It is measured in volts and can also be thought of as the 'voltage' across a component. In the context of our exercise, the potential difference between points
\( b \)
and
\( c \)
is found by multiplying the current
\( I(t) \)
at a given time by the resistance
\( R \)
with the formula
\( V_{bc} = I(t)R \)
. The sign of this voltage tells us which point has a higher potential; if positive, point
\( b \)
is at a higher potential than point
\( c \)
. This concept is directly connected to both Ohm's Law and the behavior of the circuit over time since it combines the principles of the current's magnitude at a specific instance and the resistance through which it flows.

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Most popular questions from this chapter

I-C Oscillations. A capacitor with capacitance \(6.00 \times 10^{-5} \mathrm{~F}\) is charged by connecting it to a \(12.0 \mathrm{~V}\) battery. The capacitor is disconnected from the battery and connected across an inductor with \(L=1.50 \mathrm{H}\). (a) What are the angular frequency \(\omega\) of the electrical os cillations and the period of these ascillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) Ilow much energy is initially stored in the capacitor? (d) What is the charge on the capacitor \(0.0230 \mathrm{~s}\) after the connection to the inductor is made? Interpret the sign of your answer. (c) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

A solcnoid \(25.0 \mathrm{~cm}\) long and with a cross-sectional area of \(0.500 \mathrm{~cm}^{2}\) contains 400 tums of wirc and carrics a currcnt of \(80.0 \mathrm{~A}\). Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic field if the solenoid is filled with air; (c) the total energy contained in the coil's magnetic field (assume the field is uniform); (d) the inductance of the solenoid.

A long. straight solenoid has 800 turns. When the current in the solenoid is \(2.90 \mathrm{~A}\), the average flux through each turn of the solenoid is \(3.25 \times 10^{-3} \mathrm{~Wb}\). What must be the magnitude of the rate of change of the current in order for the sclf-induced cmf to cqual \(6.20 \mathrm{mV} ?\)

If part of the magnet develops resistance and liquid helium boils away, rendering more and more of the magnct nonsuperconducting. how will this quench affect the time for the current to drop to half of its initial value? (a) The time will be shorter bccause the resistance will increase; (b) the time will be longer because the resistance will increase: (c) the time will be the same; (d) not enough information is given.

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1 ). The inner solenoid is \(25.0 \mathrm{~cm}\) long and has a diameter of \(2.00 \mathrm{~cm}\). At a certain time, the current in the inner solenoid is \(0.120 \mathrm{~A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{~A} / \mathrm{s}\). For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.
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