then use P = IV, I 2 R , or V 2 /R to find power. On the other hand, there’s a quick way to reason through
this one. Voltage changes across the 100 Ω resistor, then again across the parallel combination.
Because the 100 Ω resistor has a bigger resistance than the parallel combination, the voltage across it
is larger as well. Now consider each resistor individually. By power = V 2 /R , the 100 Ω resistor has
both the biggest voltage and the smallest resistance, giving it the most power.2 . A— The energy stored by a capacitor is ½CV 2 . By powering a car, this electrical energy is
converted into mechanical work, equal to force times parallel displacement. Solve for displacement,
you get 36 m.
3 . C —To use Ohm’s law here, simplify the circuit to a 10-V battery with the 10 Ω equivalent resistance.
We can use Ohm’s law for the entire circuit to find that 1.0 A is the total current. Because all the
resistors are in series, this 1.0 A flows through each resistor, including the 2 Ω resistor.
4 . D —First, simplify the circuit to find the equivalent capacitance. The parallel capacitors add to 6 μF.
Then the two series capacitors combine to 3 μF. So we end up with 9 V across a 3 μF equivalent
capacitance. By the basic equation for capacitors, Q = CV , the charge stored on these capacitors is 27
μC.
5 . A —An ammeter is placed in series with other circuit components. In order for the ammeter not to
itself resist current and change the total current in the circuit, you want the ammeter to have as little
resistance as possible—in the ideal case, zero resistance. But a voltmeter is placed in parallel with
other circuit components. If the voltmeter has a low resistance, then current will flow through the
voltmeter instead of through the rest of the circuit. Therefore, you want it to have as high a resistance as
possible, so the voltmeter won’t affect the circuit being measured.6 . (a) Combine each of the sets of parallel resistors first. You get 120 Ω for the first set, 222 Ω for the
second set, as shown in the diagram below. These two equivalent resistances add as series
resistors to get a total resistance of 342 Ω.
(b) Now that we’ve found the total resistance and we were given the total voltage, just use Ohm’s law
to find the total current to be 0.026 A (also known as 26 mA).
(c) and (d) should be solved together using a V-I-R chart. Start by going back one step to when we
began to simplify the circuit: a 9-V battery, a 120 Ω combination, and a 222 Ω combination,
shown above. The 26-mA current flows through each of these ... so use V = IR to get the voltage
of each: 3.1 V and 5.8 V, respectively.