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Problem 2: A lightbulb is in series with an R = 2.00 resistor. The lightbulb dissipates P = 10 W when this series circuit is
Problem 2: A lightbulb is in series with an R = 2.00 resistor. The lightbulb dissipates P = 10 W when this series circuit is con- nected to a AVia = 9.0 V battery. What is the current through the lightbulb? a) In Fig.2, draw the diagram that represents this electric cir- cuit. A library of basic symbols used for electric circuit drawings is given in paragraph 28.1 of the textbook. b) Express the resistance of the lightbulb, En, in terms of the FIG. 2: The scheme for I'roblem 2 power that it dissipates, P, and the current in the circuit, I. Do not plug in the numerical values yet, just write down the formula. c) Write down the Kirchhoff's loop law for this circuit. This should result in a quadratic equation for the current I. Formulate this equation in a symbolic form (the coefficients must be some functions of AVis, R and P). d) Plug in the numerical values of the parameters and solve the quadratic equation for I. You will get two possible values for the current. e) Naturally, the two values of / that you got in d) correspond to two different lightbulbs (a single lughtbulb of a fixed resistance uniquely determines the current in the circuit). What would be the two possible values of the rating (nominal) power of the lightbulbs that satisfy the conditions of the
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