phy1020.DVI

this currents splits at the fork with the two parallel resistors into two currents,I 2 +I 3. After passing through
these two resistors, the currents recombine, so currentI 1 runs through the bottom leg of the circuit before
returning to the battery at theterminal.
Our job is to find the currents through, and voltages across, each of the three resistors. To begin, we’ll
reduce the three resistors to a single equivalent resistor. The two resistors in parallel are equivalent to a single
resistance of

R 23 D

1

1

R 2 C

1

R 3

(22.1)

D

1

1

5C

1

3

(22.2)

D1:875  (22.3)

Now we’ve reduced the circuit to two resistors:R 1 in series withR 23. The equivalent resistance of these
two in series is

R 123 DR 1 CR 23 (22.4)

D10 C1:875  (22.5)

D11:875  (22.6)

So now we’ve reduced all three original resistors to a single equivalent resistance of9:875 connected
to the 6 V battery. We can find the currentI 1 coming out of the battery using Ohm’s law:

I 1 D

V

R 123

(22.7)

D

6 V

11:875 

(22.8)

D0:50526A (22.9)

This is also the current through resistorR 1. Since we knowR 1 and the current throughR 1 , we can use Ohm’s
law to find the potential difference acrossR 1 :

V 1 DI 1 R 1 (22.10)

D.0:50526A/.10 / (22.11)

D5:0526V: (22.12)

Now we need to find the currents through and voltages across resistorsR 2 andR 3. There are a few ways
we could proceed:

1. We can consider the two resistorsR 2 andR 3 as equivalent to a single resistorR 23 D1:875 ,as
   we’ve already worked out. The current through this equivalent resistor isI 1 D0:50526A. Knowing
   the resistance and current, we can find the voltage across the two parallel resistors. Knowing the voltage
   across each resistor, you can now work out the individual currents in each resistor using Ohm’s law.


2. Alternatively, we can note that the sum of the voltage drops for all the resistors must equal the voltage
   rise due to the battery. Therefore, the voltage drop across the two parallel resistors must be 6 V
   5:0526VD0:9474V. Knowing this voltage and the resistancesR 2 andR 3 , we can use Ohm’s law to
   solve for the currents.