Mastering Physics Solutions: Finding Current by Changing Resistors

Mastering Physics Solutions: Finding Current by Changing Resistors

On March 7, 2013, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Part A = 1.40 A Click to use the calculator/solver for this part of the problem

If the external resistance is then changed to R2 = 4.00 Ω, what is the value of the current I2 in the circuit?

Click for More...

Tagged with:  

Mastering Physics Solutions: Kirchhoff’s Current Rule Ranking Task

Mastering Physics Solutions: Kirchhoff’s Current Rule Ranking Task

On March 6, 2013, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Part A = A > (B = C)
Part B = C > (A = B)
Part C = D > A > (B = C)

Rank the resistors in the figure below (A to C) on the basis of the current that flows through them.
Rank the resistors in the figure below (A to C) on the basis of the current that flows through them.
Rank the resistors in the figure below (A to D) on the basis of the current that flows through them.

Click for More...

Tagged with:  

Mastering Physics Solutions: Exercise 18.50

Mastering Physics Solutions: Exercise 18.50

On May 5, 2012, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Part A = 2.3 A
Part B = yes
Part C = not enough
Part D = You need at least 5-10 amps to trip a circuit (think fuses). Less than an amp can kill you.

How much current is in you during the time you are touching the probes, assuming that the outlet was actually operating properly and there is a resistance of 105 Ω between your hands?
Is this enough current to be dangerous to you?
Is there enough current to “trip” the circuit breaker and save the day?
Comment on the relative sizes of your answers to parts B and C.

Click for More...

Tagged with:  

Mastering Physics Solutions: ± Capacitor Supplies Current to Bulb

Mastering Physics Solutions: ± Capacitor Supplies Current to Bulb

On May 5, 2012, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Part A = Individual charges flow through the circuit from the negative to the positive plate of the capacitor., Current flows clockwise through the circuit.
Part B = IP = IB = IN
Part C = decreases.
Part D = 6 s

Click for More...

Tagged with:  

Mastering Physics Solutions: Throw the Switch

Mastering Physics Solutions: Throw the Switch

On May 5, 2012, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Part A = The potential difference across B drops to zero, The potential difference across A increases by 50%.
Part B = The potential difference across A increases.

In this problem ε denotes the emf provided by the source, and R is the resistance of each bulb.

Click for More...

Tagged with:  

Mastering Physics Solutions: Battery, Ammeter, and Resistors

Mastering Physics Solutions: Battery, Ammeter, and Resistors

On April 20, 2012, in Chapter 18: Basic Electric Circuits, by Mastering Physics Solutions

Battery, Ammeter, and Resistors Part A = 0.667 Solution Below: An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru . The ammeter reads a current I0. Next, a resistor of unknown resistance Rr is connected in series to the ammeter, and the ammeter’s reading drops [...]

Click for More...

Tagged with: