Using Ohm’s law, determine the current in each resistor. Compare your results to the loop...

Using Ohm’s law, determine the current in each resistor. Compare your results to the loop current method in Example.

Example

For the Wheatstone bridge circuit in Figure, find the loop currents. Use the loop currents to solve for the current in each resistor (branch current).

Assign three clockwise loop currents (IA, IB, and IC) as shown in Figure Then write the loop equations. The equations for the loops are

Loop A: −12 + 330(IA – IB) + 300(IA – IC) = 0

Loop B: 330(IB −IA) + 360IB + 1000(IB –IC) = 0

Loop C: 300(IC −  IA) + 1000(IC – IB) + 390 IC = 0

Rearrange the equations into standard form:

Loop A: 630 IA − 330 IB − 300 IC = 12 V

Loop B: −330 IA + 1690 IB− 1000 IC = 0

Loop C: −300 IA −1000 IB + 1690 IC = 0

You can solve these equations with substitution, but this is tedious with three unknowns. The determinant method or directly solving with your calculator are simpler ways. Units are not shown until the end of the problem.

Evaluating the characteristic determinant using the expansion method,

= [(630)( 1690)(1690) + (−330)( −1000)( −300) + (−300)( −330)( −1000)] − [(−300)(1690)( −300) + (−1000) –(1000)(630) + (1690)( −330)( −330)] = 635202000

Solving for IA:

Solving for IB:

Solving for IC:

The current in R1 is the difference between IA and IB:

I1 = (IA− IB) = 35.1 mA − 16.2 mA = 18.9 mA

The current in R2 is the difference between IA and IC.

I2 = (IA– IC) = 35.1 mA − 15.8 mA = 19.3 mA

The current in R3 is IB:

I3 = IB = 16.2 mA

The currentin R4 is IC.

I4 = Ic = 15.8 mA

The current in RL is the difference between IB and IC:

IL = (IB – IC) = 16.2 mA − 15.8 mA = 0.4 mA