A real battery is not just an emf. We can If 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 Figure P23.55. 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. the value of the emf. Suppose the terminals of this battery are connected to a 2.0 Ω resistor. Figure P23.55 a. What is the potential difference between the terminals of the battery? b. What fraction of the battery’s power is dissipated by the internal resistance?
A real battery is not just an emf. We can If 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 Figure P23.55. 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. the value of the emf. Suppose the terminals of this battery are connected to a 2.0 Ω resistor. Figure P23.55 a. What is the potential difference between the terminals of the battery? b. What fraction of the battery’s power is dissipated by the internal resistance?
A real battery is not just an emf. We can If 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 Figure P23.55. 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. the value of the emf. Suppose the terminals of this battery are connected to a 2.0 Ω resistor.
Figure P23.55
a. What is the potential difference between the terminals of the battery?
b. What fraction of the battery’s power is dissipated by the internal resistance?
Use the following information to answer the next question.
Two mirrors meet an angle, a, of 105°. A ray of light is incident upon mirror A at an angle, i, of
42°. The ray of light reflects off mirror B and then enters water, as shown below:
A
Incident
ray at A
Note: This diagram is not to
scale.
Air (n = 1.00)
Water (n = 1.34)
B
Use the following information to answer the next question.
Two mirrors meet an angle, a, of 105°. A ray of light is incident upon mirror A at an angle, i, of
42°. The ray of light reflects off mirror B and then enters water, as shown below:
A
Incident
ray at A
Note: This diagram is not to
scale.
Air (n = 1.00)
Water (n = 1.34)
B
Good explanation it sure experts solve it.
Chapter 23 Solutions
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