PHYS208 Fundamentals of Physics II
Quiz 2 - dc Circuits
Two lightbulbs are connected in series with the battery shown.
- What is the voltage difference between a
and b? (That is, what is the terminal voltage of the
- A third identical lightbulb is now connected in parallel to one of the other
By what percentage does the power dissipated by the internal resistance change?
- The equivalent resistance of the circuit above is
(2.0 + 7.0 + 7.0) ohm = 16.0 ohm.
Accordingly the current through the battery is
(9 V) / (16 ohm) = 0.56 A.
The voltage drop across the internal resistance is (0.56 A)(2.0 ohm) = 1.1 V;
the terminal voltage of the battery is thus 9.0 V - 1.1 V = 7.9 V.
- The additional lightbulb is added as shown below:
The two 7.0 ohm resistors in parallel are equivalent to a 3.5 ohm resistance;
the total equivalent resistance "seen" by the emf is thus
(2.0 + 3.5 + 7.0) ohm = 12.5 ohm.
Accordingly the current through the battery is now
(9 V) / (12.5 ohm) = 0.72 A.
Since the power dissipated by a resistor is proportional to the square of the current,
the power ratio of the internal resistance will be
(i2/i1)2 = (0.72/0.56)2 = 1.65.
The power dissipated by the internal resistance increases by 65% when the third lightbulb
Last updated Feb. 27, 1998.
Copyright George Watson, Univ. of Delaware, 1998.