This is my circuit. I need to find the power dissipated by Vlamp. So the lamp turns on if the Vlamp is more than 2V. When the lamp is on it has an internal resistance of 10 Ohms, and when it is off it acts like an open circuit.
How would I find the power dissipated? I know how to do it for Voltage sources and Resistors, but not elements like these.
One more thing, I already figured out that:Battery has open circuit voltage of 4VR1 = 5 OhmsR2 = 5 Ohmsk = 0.2A/VWhen lamp on and Vlamp = 2V, Rbat = 10 Ohms
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Answer
Use the power equations, P=I^2*R and P=V^2/R, you can use either one.
For the lamp, if its on, the equation would be \$P = V_{lamp}^2/2\$
You can analyze the circuit like normal to find V_lamp
Another way you could do this is if you find the current and the voltage through the lamp, then use the equation P=I*V
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