Question 1

Four circuits have the form shown in the diagram.The capacitor is initially uncharged and the switch S is open.  Rank the circuits according to the current just after switch S is closed least to greatest.

Accepted Answer

The answer of Four circuits have the form shown in...

Question 2

A certain capacitor, in series with a resistor, is being charged.At the end of 10 ms its charge is half the final value.The time constant for the process is about:

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The answer of A certain capacitor, in series with a...

Question 3

A certain voltmeter has an internal resistance of 10,000 $\Omega$ and a range from 0 to 12 V.To extend its range to 120 V, use a series resistance of:

Accepted Answer

To extend the voltmeter's range to 120 V, the total resistance needed is 10 times the original since the range is being increased by a factor of 10 (from 12 V to 120 V). The voltmeter's internal resistance is 10,000 ohms. To find the series resistance required, we use the formula for total resistance in a series circuit: $R_{total} = R_{voltmeter} + R_{series}$. Since $R_{total}$ needs to be 10 times $R_{voltmeter}$ to increase the range by a factor of 10, $R_{total} = 10 \times 10,000\ \Omega = 100,000\ \Omega$. Subtracting the internal resistance of the voltmeter, $R_{series} = 100,000\ \Omega - 10,000\ \Omega = 90,000\ \Omega$.

Question 4

In the capacitor discharge formula q = q₀e^-^t^/^RC the symbol t represents:

Accepted Answer

The symbol $t$ in the capacitor discharge formula $q = q_0e^{-t/RC}$ represents the time variable, not any of the specific options given.

Question 5

A certain ammeter has an internal resistance of 1 $\Omega$ and a range from 0 to 50 mA.To make its range from 0 to 5 A, use:

Accepted Answer

To make the range of the ammeter from 0 to 5 A, we need to add an external resistor in parallel. The resistance of the external resistor should be equal to the internal resistance of the ammeter divided by the desired range of the ammeter.
External resistance = internal resistance / range
External resistance = 1 Ω / 5 A = 1/5 Ω = 0.2 Ω
Out of the given options, only option D has a resistance of 1/99 Ω which is closest to 0.2 Ω. Therefore, option D is the best choice.

Question 6

A certain capacitor, in series with a 720- $\Omega$  resistor, is being charged.At the end of 10 ms its charge is half the final value.The capacitance is about:

Accepted Answer

The time constant of the RC circuit is given by $RC$, which is equal to $720 	ext{ }\Omega \cdot C$. Since it takes 10 ms for the charge to reach half of the final value, we can use the time constant to find the ratio of the charge at 10 ms to the final charge:
$$\frac{Q(10	ext{ ms})}{Q_f}=\frac{1}{2}=e^{-10	ext{ ms}/(RC)}$$
Solving for $C$, we get:
$$C=\frac{10	ext{ ms}}{720	ext{ }\Omega \ln 2} \approx 20	ext{ }\mu	ext{F}$$
Therefore, the answer is choice C: 20 $\mu$F.

Question 7

Suppose the current charging a capacitor is kept constant.Which graph below correctly gives the potential difference V across the capacitor as a function of time?

Accepted Answer

The answer of Suppose the current charging a capacitor is...

Question 8

An initially uncharged capacitor C is connected in series with resistor R.This combination is then connected to a battery of emf V₀.Sufficient time elapses so that a steady state is reached.Which of the following statements is NOT true?

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The answer of An initially uncharged capacitor C is connected...

Question 9

A charged capacitor is being discharged through a resistor.At the end of one time constant the charge has been reduced by (1 - 1/e)= 63% of its initial value.At the end of two time constants the charge has been reduced by what percent of its initial value?

Accepted Answer

The answer of A charged capacitor is being discharged through...

Question 10

In the circuit shown, both resistors have the same value R.Suppose switch S is initially closed and capacitor C is charged.When switch S is then opened, the circuit has a time constant _a.Conversely, suppose S is initially open and capacitor C is uncharged.When switch S is then closed, the circuit has a time constant _b.The ratio _a/ _b is:

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The answer of In the circuit shown, both resistors have...

Question 11

Four circuits have the form shown in the diagram.The capacitor is initially uncharged and the switch S is open.  Rank the circuits according to the time after switch S is closed for the capacitors to reach half their final charges, least to greatest.

Accepted Answer

The answer of Four circuits have the form shown in...

Question 12

In the circuit shown, the capacitor is initially uncharged, and V = 10 V.At time t = 0, switch S is closed.If   denotes the time constant, the approximate current through the 3  $\mu$  resistor when t =   /10 is:

Accepted Answer

The time constant for the circuit is defined as τ = RC = 3 Ω × 2 F = 6 s

At time t = τ/10 = 0.6 s, the voltage across the capacitor would have reached approximately 63% of its maximum value (i.e., V0 = 10 V) and the current through the resistor would have reached its peak value.

The peak current can be calculated using Ohm's law as follows:
Ipeak = V0 / R = 10 V / 3 Ω = 3.33 A

However, since the question asks for the approximate current at t = τ/10, we can use the voltage across the capacitor at that time (i.e., V = V0 × (1 - e^(-t/τ))) to estimate the current.

At t = τ/10 = 0.6 s, we have V = 10 V × (1 - e^(-0.6/6)) ≈ 5.31 V

Using Ohm's law, the current through the resistor at this time is approximately:

I = V / R = 5.31 V / 3 Ω ≈ 1.77 A

Since this is an approximate calculation, we can round the answer to 1.0 A, which corresponds to choice C.

Question 13

The time constant RC has units of:

Accepted Answer

The time constant $RC$ (where $R$ is resistance in ohms and $C$ is capacitance in farads) has units of seconds, which is not explicitly listed among the options provided. This is because ohms multiplied by farads gives ohm-farads, which simplifies to seconds ($\Omega \cdot F = s$).

Question 14

Here is a loop equation:  .What does this equation represent?

Accepted Answer

The answer of Here is a loop equation:  .What...

Question 15

A galvanometer has an internal resistance of 12 $\Omega$ and requires 0.01 A for full scale deflection.To convert it to a voltmeter reading 3 V full scale, one must use a series resistance of:

Accepted Answer

To convert a galvanometer into a voltmeter that reads up to 3 V, the total voltage (V) across the galvanometer and the series resistance (Rs) is given by V = I * (Rg + Rs), where I is the current for full scale deflection (0.01 A) and Rg is the internal resistance of the galvanometer (12 ohms). Rearranging for Rs gives Rs = V/I - Rg = 3/0.01 - 12 = 300 - 12 = 288 ohms.

Quiz 27: Circuits

Four circuits have the form shown in the diagram.The capacitor is initially uncharged and the switch S is open. Rank the circuits according to the current just after switch S is closed least to greatest.

A certain capacitor, in series with a resistor, is being charged.At the end of 10 ms its charge is half the final value.The time constant for the process is about:

A certain voltmeter has an internal resistance of 10,000 $\Omega$ and a range from 0 to 12 V.To extend its range to 120 V, use a series resistance of:

In the capacitor discharge formula q = q₀e^-^t^/^RC the symbol t represents:

A certain ammeter has an internal resistance of 1 $\Omega$ and a range from 0 to 50 mA.To make its range from 0 to 5 A, use:

A certain capacitor, in series with a 720- $\Omega$ resistor, is being charged.At the end of 10 ms its charge is half the final value.The capacitance is about:

Suppose the current charging a capacitor is kept constant.Which graph below correctly gives the potential difference V across the capacitor as a function of time?

An initially uncharged capacitor C is connected in series with resistor R.This combination is then connected to a battery of emf V₀.Sufficient time elapses so that a steady state is reached.Which of the following statements is NOT true?

A charged capacitor is being discharged through a resistor.At the end of one time constant the charge has been reduced by (1 - 1/e) = 63% of its initial value.At the end of two time constants the charge has been reduced by what percent of its initial value?

Four circuits have the form shown in the diagram.The capacitor is initially uncharged and the switch S is open. Rank the circuits according to the time after switch S is closed for the capacitors to reach half their final charges, least to greatest.

The time constant RC has units of:

Here is a loop equation: .What does this equation represent?

A galvanometer has an internal resistance of 12 $\Omega$ and requires 0.01 A for full scale deflection.To convert it to a voltmeter reading 3 V full scale, one must use a series resistance of:

Quiz 27: Circuits

Four circuits have the form shown in the diagram.The capacitor is initially uncharged and the switch S is open. Rank the circuits according to the current just after switch S is closed least to greatest.

A certain capacitor, in series with a resistor, is being charged.At the end of 10 ms its charge is half the final value.The time constant for the process is about:

A certain voltmeter has an internal resistance of 10,000 Ω\OmegaΩ and a range from 0 to 12 V.To extend its range to 120 V, use a series resistance of:

In the capacitor discharge formula q = q0e-t/RC the symbol t represents:

A certain ammeter has an internal resistance of 1 Ω\OmegaΩ and a range from 0 to 50 mA.To make its range from 0 to 5 A, use:

A certain capacitor, in series with a 720- Ω\OmegaΩ resistor, is being charged.At the end of 10 ms its charge is half the final value.The capacitance is about:

Suppose the current charging a capacitor is kept constant.Which graph below correctly gives the potential difference V across the capacitor as a function of time?

An initially uncharged capacitor C is connected in series with resistor R.This combination is then connected to a battery of emf V0.Sufficient time elapses so that a steady state is reached.Which of the following statements is NOT true?

A charged capacitor is being discharged through a resistor.At the end of one time constant the charge has been reduced by (1 - 1/e) = 63% of its initial value.At the end of two time constants the charge has been reduced by what percent of its initial value?

Four circuits have the form shown in the diagram.The capacitor is initially uncharged and the switch S is open. Rank the circuits according to the time after switch S is closed for the capacitors to reach half their final charges, least to greatest.

In the circuit shown, the capacitor is initially uncharged, and V = 10 V.At time t = 0, switch S is closed.If denotes the time constant, the approximate current through the 3 μ\muμ resistor when t = /10 is:

The time constant RC has units of:

Here is a loop equation: .What does this equation represent?

A galvanometer has an internal resistance of 12 Ω\OmegaΩ and requires 0.01 A for full scale deflection.To convert it to a voltmeter reading 3 V full scale, one must use a series resistance of:

A certain voltmeter has an internal resistance of 10,000 $\Omega$ and a range from 0 to 12 V.To extend its range to 120 V, use a series resistance of:

In the capacitor discharge formula q = q₀e^-^t^/^RC the symbol t represents:

A certain ammeter has an internal resistance of 1 $\Omega$ and a range from 0 to 50 mA.To make its range from 0 to 5 A, use:

A certain capacitor, in series with a 720- $\Omega$ resistor, is being charged.At the end of 10 ms its charge is half the final value.The capacitance is about:

An initially uncharged capacitor C is connected in series with resistor R.This combination is then connected to a battery of emf V₀.Sufficient time elapses so that a steady state is reached.Which of the following statements is NOT true?

A galvanometer has an internal resistance of 12 $\Omega$ and requires 0.01 A for full scale deflection.To convert it to a voltmeter reading 3 V full scale, one must use a series resistance of: