4. When a light bulb is connected to the simple circuit shown, it dissipates a p

Question

4. When a light bulb is connected to the simple circuit shown, it dissipates a power of 60.0 W. (a) Determine the possible values for the current in the circuit. (b) Suppose that the values of the resistor's resistance is varied without changing the bulb's rate of energy dissipation. What is the maximum resistance? (c) If instead the battery's terminal potential is varied without affecting the bulb's power, what is the minimum terminal potential? BULB-, non-oh 4.00 ? 32.0 V No n - Ohais If ohnic

Explanation / Answer

a) Power is given by P = V I    here V is voltage and I is current

                  V = 32 volt( as given in part c ) ; I = P/ V

              I = 60 / 32 = 1.875 Amperes

b) without changing the power, maximum resistance is given as

                                    P = I2 Rmax

                                 Rmax = 60 / (1.8752 ) = 17.066 ohm

c) for finding terminal potential difference;

                          V = E - Ir   

    V= terminal potential difference ; E = emf (potential of open circuit) ; I = current ; r = internal resistance of bulb

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