8.1

As discussed above, the first step is to simplify the circuit by replacing the two parallel resistors with a single resistor that has an equivalent resistance. Two 8 Ω resistors in series is equivalent to a single 4 Ω resistor. Thus, the two branch resistors (R2 and R3) can be replaced by a single resistor with a resistance of 4 Ω. This 4 Ω resistor is in series with R1 and R4. Thus, the total resistance is

Rtot = R1 + 4 Ω + R4 = 5 Ω + 4 Ω + 6 Ω

Rtot = 15 Ω

Now the Ohm's law equation (ΔV = I • R) can be used to determine the total current in the circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be used.

Itot = ΔVtot / Rtot = (60 V) / (15 Ω)

Itot = 4 Amp

The 4 Amp current calculation represents the current at the battery location. Yet, resistors R1 and R4 are in series and the current in series-connected resistors is everywhere the same. Thus,

Itot = I1 = I4 = 4 Amp

For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I2 + I3 must equal 4 Amp. There are an infinite number of possible values of I2 and I3 that satisfy this equation. Since the resistance values are equal, the current values in these two resistors are also equal. Therefore, the current in resistors 2 and 3 are both equal to 2 Amp.

I2 = I3 = 2 Amp

Now that the current at each individual resistor location is known, the Ohm's law equation (ΔV = I·R) can be used to determine the voltage drop across each resistor. These calculations are shown below.

ΔV1 = I1 ·R1 = (4 Amp) ·(5 Ω)

ΔV1 = 20 V

ΔV2 = I2 · R2 = (2 Amp) ·(8 Ω)

ΔV2 = 16 V

ΔV3 = I3 · R3 = (2 Amp) · (8 Ω)

ΔV3 = 16 V

ΔV4 = I4 ·R4 = (4 Amp) · (6 Ω)

ΔV4 = 24 V

The analysis is now complete and the results are summarized in the diagram below.