In Figure (a), both batteries have emf ε =
1.70 V and the external resistance R is a variable
resistor. Figure (b) gives the electric potentials
V between the terminals of each battery as functions of
R: Curve 1 corresponds to battery 1, and curve 2
corresponds to battery 2. The horizontal scale is set by
Rs = 0.200 Ω. What is the internal
resistance of (a) battery 1 and
(b) battery 2?
In Figure (a), both batteries have emf ε = 1.70 V and the external resistance R...
In Figure (a), both batteries have emf ε = 1.30 V and the
external resistance R is a variable resistor. Figure (b) gives the
electric potentials V between the terminals of each battery as
functions of R: Curve 1 corresponds to battery 1, and curve 2
corresponds to battery 2. The horizontal scale is set by Rs = 0.300
Ω. What is the internal resistance of (a) battery 1 and (b) battery
2?
0.65 R 0 R 0.390 R (22)...
You have two batteries, both with an emf of 1.5V. The internal resistance of the first battery is 0.5 Ohm, and the internal resistance of the second battery is 1 Ohm What will be the current through an external resistor with a resistance of 2 Ohm, if the two batteries are connected in series? What will be the current through an external resistor with a resistance of 2 Ohm, if the two batteries are connected in parallel? What will be...
You have two batteries, both with an emf of 1.5V. The internal resistance of the first battery is 0.5 Ohm, and the internal resistance of the second battery is 1 Ohm. What will be the current through an external resistor with a resistance of 2 Ohm, if the two batteries are connected in series? What will be the current through an external resistor with a resistance of 2 Ohm, if the two batteries are connected in parallel? What will be...
A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure(Figure 1) . A typical battery has 1.0 Ω internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V,...
R E, In the figure , battery 1 has emf E1 = 12.0V and internal resistance rı = 0.016 Ohms and battery 2 has emf E2 = 12.0V and internal resistance r2 = 0.014 Ohms. The batteries are connected in series with an external resistance R. What R value in ohms makes the terminal-to- terminal potential difference of one of the batteries zero?
A battery with an emf of 3.00 V has an internal resistance r. When connected to a resistor R, the terminal voltage is 2.90 V and the current is 0.16 A. 1. What is the value of the external resistor R? 2. What is the internal resistance r of the battery? 3. What is the energy dissipated in the battery's internal resistance in 1.6 minutes? 4. When a second identical battery is added in series and the external resistor is...
A battery has an emf of 12.0 V and an internal resistance of 0.210 Q. Its terminals are connected to a load resistance of 3.00 . Circuit diagram of a source of emf (in this case, a battery), of internal resistance r, connected to an external resistor of resistance R. for ning R (a) Find the current in the circuit and the terminal voltage of the battery. SOLUTION Conceptualize Study the figure, which shows a circuit consistent with the problem...
A resistor with resistance R is connected to a battery that has emf 14.0 V and internal resistance r = 0.410 Ω. For what two values of R will the power dissipated in the resistor be 81.0W? (There is two answers)
2. The following figure both batteries have negligible internal
resistance and the ammeter reads a stream of 2.50 in the sense that
it illustrates. Find EMF of the battery polarity indicated is
correct?
12.0 Ω 上 48.0 15.0 (2 75.0 V