## PHYS345 Electricity and Electronics

Find the current in each resistor and the voltage difference between points
*a* and *b*.
Let emf_{1} = 6.0 V,
emf_{2} = 5.0 V,
emf_{3} = 4.0 V,
R_{1} = 100 ohm,
and R_{2} = 50 ohm.

Also, what is the current through emf_{2}?

Which resistor is dissipating more power?

### Solution:

First I would recommend redrawing the circuit to a configuration with which you may be more comfortable. This is an attempt to show which elements are in parallel (and which are not).

Although the full approach relying on solution of simultaneous equations is not required,
it is not a bad idea to follow some of the suggestions for solving multiloop equations.
That is, pick current directions and label circuit accordingly.

By inspection, we notice that R_{1} is in parallel
with a "perfect" emf_{2}.
Ohm's law may thus be readily applied to yield the current in
R_{1} of emf_{2} / R_{1}
= 5 V / 100 ohm = 50 mA.

For the current in R_{2}, it is best to apply the loop rule.
From that, you would find that the effective emf experienced by R_{2}
is emf_{2} - emf_{1} + emf_{3}.
Thus the current in R_{2} is (5-6+4 V) / 50 ohm = 3 V / 50 ohm = 60 mA.

The voltage difference between *a* and *b* is best seen from the original figure
to be emf_{3} + emf_{2} = 9.0 V.

Applying the junction rule to either of the two junctions shows that the current
in emf_{2} is the sum of the current in R_{1} and R_{2} which is 110 mA.

The power dissipated by R_{1} is (50 mA)(5 V) = 250 mW.

The power dissipated by R_{2} is (60 mA)(3 V) = 180 mW.

R_{1} is dissipating more power than R_{2}.

When time permits, you may check the current/voltage solutions of cicuits by
confirming the energy conservation;
*i.e.*, make sure that the power delivered by emfs equals the power dissipated
by all resistors plus the energy absorbed by any emfs that have current flowing
against their polarity (recharging).
From the answer to the final question on the quiz,
we have that 430 mW is being dissipated by the two resistors.
The powers associated with the three emfs are:

emf_{1} |
(110 mA)(5 V) |
= 550 mW |
delivered |

emf_{2} |
(60 mA)(6 V) |
= 360 mW |
absorbed |

emf_{3} |
(60 mA)(4 V) |
= 240 mW |
delivered |

total |
(550 - 360 + 240) mW |
= 430 mW |
as expected |

"http://www.physics.udel.edu/~watson/phys345/examples/multiloop-example.html"

Last updated Sept. 7, 1998.

Copyright George Watson, Univ. of Delaware, 1998.