PHYS345 Electricity and Electronics

Find the current in each resistor and the voltage difference between points a and b. Let emf1 = 6.0 V, emf2 = 5.0 V, emf3 = 4.0 V, R1 = 100 ohm, and R2 = 50 ohm.

multiloop circuit

Also, what is the current through emf2?

Which resistor is dissipating more power?


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).

multiloop circuit

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.

multiloop circuit

By inspection, we notice that R1 is in parallel with a "perfect" emf2. Ohm's law may thus be readily applied to yield the current in R1 of emf2 / R1 = 5 V / 100 ohm = 50 mA.

For the current in R2, it is best to apply the loop rule. From that, you would find that the effective emf experienced by R2 is emf2 - emf1 + emf3. Thus the current in R2 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 emf3 + emf2 = 9.0 V.

Applying the junction rule to either of the two junctions shows that the current in emf2 is the sum of the current in R1 and R2 which is 110 mA.

The power dissipated by R1 is (50 mA)(5 V) = 250 mW.
The power dissipated by R2 is (60 mA)(3 V) = 180 mW.
R1 is dissipating more power than R2.

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:

emf1 (110 mA)(5 V)  = 550 mW delivered
emf2 (60 mA)(6 V)  = 360 mW absorbed
emf3 (60 mA)(4 V)  = 240 mW delivered
total   (550 - 360 + 240) mW  = 430 mW as expected

Last updated Sept. 7, 1998.
Copyright George Watson, Univ. of Delaware, 1998.