Answered: 5. In the circuit on the right, the emf… | bartleby

Related questions

Question

5. In the circuit on the right, the emf (voltage) of the power
supply is V = 120V. Resistors R₁ = 5.0, R₂ = 20.00, R3 =
30.0 0.
-
(a) Calculate the current through R₁ before and after the v
switch S is closed ("closed" means "connected").
(b) Calculate the electric power dissipated by R₁ before and
after the switch S is closed.
R₁
S
R₂
R3.

Expert Solution

Step by step

Solved in 5 steps with 19 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

. A resistor operated at 120 V DC is rated at 1.40 kW. a) What is the normal operating current through the resistor? b) What is the resistance of the resistor?
In the circuit shown below, R1=10 N, R2=32 Q. Ω. Find the equivalent resistance of this circuit: Reg = A. Find the current passing through the R1 resistor: I1= A. Find the current passing through the 5-2 resistor: 15= W. Find the power dissipated on the 5-0 resistor: P5= 10 Q 50 60 V + 10 Q R2 R1 R2
5. An electric circuit contains a resistance R and a capacitor C in series, and a battery supplying a time-varying electromotive force (or voltage) V(t). The charge q on the capacitor therefore obeys the following differential equation: Rdq + 2 = V(t). dt Assuming that initially there is no charge on the capacitor, and given that V(t) Vosinut, where V₂ and w are some constants, find the charge on the capacitor as a function of time. = Hint: You may have to do integration by parts twice to evaluate the following inte- gral: feRc sin(wt)dt.
5. This problem is about charging a weak battery & = 10.8 V with a good battery &, = 12.0 V. Let the internal battery resistances be r₁ = r₂ = 0.022, and resistance of the charging cable be R = 0.02 02. Also, let the current I leave the good battery. (a) In the picture below, the battery is hooked up correctly, with the positive terminals connected to each other. Apply the loop rule to this circuit and find the current charging the battery. WH $5 R No (b) In the picture below, the battery is hooked up incorrectly. Find the current charging the battery. www R
The values of the components in a simple series RC circuit containing a switch and an initially uncharged capacitor (see figure below) are C = 1.50 µF, R = 2.30 MN, and E = 10.0 V. R (a) the charge on the capacitor... ...a long time after the switch is closed 900 = ..4.1 s after the switch is closed q = (b) the current in the resistor... ...immediately after the switch is closed In = HA ..4.1 s after the switch is closed I = HA ... a long time after the switch is closed I, = (c) The rate at which energy, 4.1 s after the switch is closed, is... ...being dissipated in the resistor Presistor = ...being stored in the capacitor Рсаpаcitor %3D µW
1F IF 6V 6 V 10 (A) (B) Figure 1:
14. A power supply providing 85 V is hooked in series to a lightbulb, using two connecting wires. The lightbulb has a resistance bulb B. of 130 N and the wires have a resistance of 15 2 each. The actual question is, how much power does the bulb use, and how much power is lost in the wires? This is broken down into steps. a. What is the equivalent resistance of the series circuit? [160 Q] b. What is the current through the circuit? [.53 A] c. What is the magnitude of the potential difference between points A and B? (What is the voltage across wire 1?) [8 V] d. What is the magnitude of the potential difference between points wire 1 wire 2 A B and C? (What is the voltage across the bulb?) [69 V – note that the voltage across the bulb is NOT the 85 V of the power supply. The sum of the voltages, 8+69+8, around the circuit, IS the voltage of the power supply.) e. How much power is used by the bulb? [36.6 W] f. How much power is lost in the one of the wires? [4.2 W] |