The following figure shows a circuit that contains two batteries and two resistors. Use Kirchhoff’s Loop Rule to determine the current in the circuit ?

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1 Answer
Apr 7, 2018

The current #I# is 0.9 A.

Explanation:

To apply Kirchoff's Loop rule we have to go (or imagine going) all the way round the loop (or loops), adding up the net voltage drops along the way. The total of the voltage drops must be zero. (In this approach, we count voltage rises as negative voltage drops.)

A quick reminder about the signs of the voltages

  • if you cross a resistor in the direction of the current, the voltage drops, and is counted with a + sign
  • if you cross a battery, the voltage will have a + sign when you go from the positive end to the negative end.

(You could, of course, keep track of the net voltage rises instead! Then each of the signs above would have reversed. Since the negative ogf zero is still zero, this will make no difference to the final equation(s).)

Let us imagine going along the loop in the circuit from A clockwise until we get back to A. We cross four circuit elements on the way :

  1. The #12\ Omega# resistor - we are crossing it in the direction of the current #I# - voltage drop is #+12\ Omega\times I#
  2. The 6.0 V battery - crossing from positive to negative - counts as a drop of #+6 # V.
  3. The #8\ Omega# resistor - we are crossing this again in the direction of the current #I# - voltage drop is #+8\ Omega\times I#
  4. The 24 V battery - this we are crossing from negative end to positive end, and hence there is actually a rise in voltage and this should count as a drop of #-24# V

Adding all these together and equating the sum to zero according to Kirchoff's loop law, we get the equation

#+12\ Omega\times I+6" V"+8\ Omega\times I-24" V"=0#

which implies

#20\ Omega\times I = 18" V"#

and thus

#I = (18" V")/(20\ Omega) = 0.9" A"#