A battery in combination with a dependent current source is connected to a lamp in the circuit shown below in Figure 6-1. The battery is modeled as a voltage source in series with a resistor, Rbat, and has an open circuit voltage of 4V. The lamp turns on if the voltage across the lamp is greater than 2V. When the lamp is on it has an internal resistance of 10Ω, and when it is off it acts like an open circuit.

asked by kanimozhi

8 years ago

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1 answer

I = V/r = 2/10 = 0.2A.

Rbat = V/I = (4-2)/0.2 = 10 Ohms.

answered by Henry

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Question

A battery in combination with a dependent current source is connected to a lamp in the circuit shown below in Figure 6-1. The battery is modeled as a voltage source in series with a resistor, Rbat, and has an open circuit voltage of 4V. The lamp turns on if the voltage across the lamp is greater than 2V. When the lamp is on it has an internal resistance of 10Ω, and when it is off it acts like an open circuit.

1 answer

I = V/r = 2/10 = 0.2A.

Rbat = V/I = (4-2)/0.2 = 10 Ohms.

Ask a New Question or answer this question.

Henry · Answer

I = V/r = 2/10 = 0.2A. Rbat = V/I = (4-2)/0.2 = 10 Ohms.