7, A constant current supply is set to deliver 250 mA. It is applied to a 15 res

Question

7, A constant current supply is set to deliver 250 mA. It is applied to a 15 resistor. What is the voltage across the resistor? 8, A real 1.5 V battery has an internal resistance of 2 . What is the maximum amount of current that one can possibly get out of the battery by connecting a heavy wire from one end to the other? How much effect does this internal resistance have if the battery is lightinga bulb which has an effective resistance at 1.5 V of 10.2? (You may make the assumption that the effective resistance of two resistors in series is just the sum of the two resistances, as you might have discovered on a previous problem set.) Express your answer as a percentage of the current that an ideal 1.5 V battery would deliver a. b. 9. A constant current supply and a constant voltage supply are both adjusted so that either one will light a particular brand of light bulb equally brightly. Now, a set of identical bulbs is put together in the arrangement below. Which power supply (without any readjustments) makes the arrangement brighter, constant current, or constant voltage? Explain c. Power Supply d. Is the current flow distribution (i.e. ratio of current through top pair of bulbs to the current through the single bottom one) dependent on which type of power supply is used?

Explanation / Answer

As the number of questions are more, I am solving the first two questions.

(7) Given that, I = 250 mA = 250 / 1000 = 0.25 A

R = 15.0 ohm

So, voltage across the resistor, V = I*R = 0.25*15 = 3.75 V

(8) (a) Current will be maximum, when the resistance of the connected wire is minimum.

at the limiting condition when resistance of the connected wire = 0

Maximum current Imax = V / r = 1.5 / 2 = 0.75 A

(b) When the battery is ideal, means its internal resistance = 0

current I1 = V / R = 1.5 / 10 = 0.15 A

In actual condition, internal resistance r = 2 ohm

so, current I2 = V / (r+R) = 1.5 / (2+10) = 0.125 A

So, the percentage change in the current = [(0.15 - 0.125) / 0.15] x 100 = 16.67 %

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