University

Physics 2

Physics 2

Fall 2024

This circuit has two resistors, with \(R_1 \gt R_2\).

- Which of the two resistors has greater current?
- Which of the two resistors has greater voltage?
- Which of the two resistors dissipates the larger amount of power?

- \(I_1 = I_2\)
- \(V_1 \gt V_2\)
- \(P_1 \gt P_2\)

This circuit has two resistors, with \(R_1 \gt R_2\).

- Which of the two resistors has greater current?
- Which of the two resistors has greater voltage?
- Which of the two resistors dissipates the larger amount of power?

- \(I_2 \gt I_1\)
- \(V_1 = V_2\)
- \(P_2 \gt P_1\)

Consider the resistor network shown.

- Resistor 8 is in parallel with which other resistor(s)?
- Resistor 1 is in parallel with which other resistor(s)?
- Resistor 7 is in parallel with which other resistor(s)?

- 6, 9, 5
- 2
- 4, 11

The figure shows five combinations of identical resistors. Rank in order, from largest to smallest, the equivalent resistances \((R_{\rm{eq}})_1\) to \((R_{\rm{eq}})_5\)

\((R_{\rm{eq}})_2 > (R_{\rm{eq}})_4 > (R_{\rm{eq}})_1 > (R_{\rm{eq}})_5 > (R_{\rm{eq}})_3\)

\((R_{\rm{eq}})_1 = R\)

\((R_{\rm{eq}})_2 = 3 R\)

\((R_{\rm{eq}})_3 = \frac{1}{3} R\)

\((R_{\rm{eq}})_4 = \frac{3}{2} R\)

\((R_{\rm{eq}})_5 = \frac{2}{3} R\)

Bulbs \(A\), \(B\) and \(C\) are identical. What would happen to all bulbs (brighter, dimmer, go out) in the following cases?

- Bulb \(C\) is cut free at points 1 and 2, leaving an open circuit.
- A wire is connected between points 1 and 2.

- \(C\) goes out, \(A\) gets dimmer, and \(B\) gets brighter
- \(A\) gets brighter, \(B\) and \(C\) go out

Reasons: Bulb brightness depends on power: \(P = I^2 R = \frac{V^2}{R}\). A bulb’s resistance won’t change, so if you can determine that either \(I\) or \(V\) increase for a bulb, then it will get brighter.

- Initially, \(A\) has more of the voltage than \(B\) (more than half of the battery voltage). Finally, \(A\) and \(B\) both have one half of the battery voltage. So \(V_A\) decreases (\(A\) gets dimmer), and \(V_B\) increases (\(B\) gets brighter).
- A wire from 1 to 2 is a zero-resistance parallel branch that will short out \(B\) and \(C\) – they both go out. Overall circuit resistance goes down, so \(A\) gets brighter.

A homemade capacitor is constructed of 2 sheets of aluminum foil with an area of \(2.00\) square meters, separated by paper, \(0.05\) mm thick, of the same area and a dielectric constant of \(3.7\). The homemade capacitor is connected in series with a \(100.00\) Ω resistor, a switch, and a \(6.00\) V voltage source.

- What is the RC time constant of the circuit?
- What is the initial current through the circuit, when the switch is closed?
- How long does it take the current to reach one third of its initial value?

- \(\displaystyle \tau = RC = R \left ( k \epsilon_0 \frac{A}{d} \right ) = 0.131\) ms
- \(\displaystyle I_0 = \frac{V}{R} = 0.06\) A
- \(t_{1/3} = (-0.131 \times 10^{-3}) \ln \frac{1}{3} = 144\) μs

A student makes a homemade resistor from a graphite pencil \(5.00\) cm long, where the graphite is \(0.05\) mm in diameter. The resistivity of the graphite is \(\rho = 1.38 \times 10^{−5}\) Ω/m. The homemade resistor is placed in series with a switch, a \(10.00\) mF capacitor and a \(0.50\) V power source.

- What is the RC time constant of the circuit?
- What is the potential drop across the pencil \(1.00\) s after the switch is closed?

- \(\tau = RC = 3.51\) s
- \(V = 376\) mV

Consider the circuit shown.

- Determine the equivalent resistance and the current from the battery with switch \(S_1\) open.
- Determine the equivalent resistance and the current from the battery with switch \(S_1\) closed.

- \(R_{\rm{eq}} = 12.0\) Ω, \(I = 1.0\) A
- \(R_{\rm{eq}} = 12.0\) Ω, \(I = 1.0\) A

When the switch is open,

*Last modified: Wed October 16 2024, 02:18 PM.*