In the circuit above, the resistors are:

R1 = 6 W

R2 = 36 W

R3 = 12 W

R4 = 3 W

What is the potential difference across each resistor? How much current passes through each resistor?

To solve this we need to find the equivalent resistance of the set of capacitors. We'll contract the circuit from 4 resistors to 1.

Step 1 - R₂ and R₃ are in parallel. Replace this pair by a single capacitor R₂₃:

1 R23

=

1 R2

+

1 R3

=

1 36

+

1 12

=

4 36

Therefore R23

=

36 4

=

9 W

Step 2 - R₄ and R₂₃ are in series. Replace that pair by a single resistor R₂₃₄ = 3 + 9 = 12 W.

Step 3 - R₁ and R₂₃₄ are in parallel. Replace that pair by a single resistor R_eq:

1 Req

=

1 R1

+

1 R234

=

1 6

+

1 12

=

3 12

Req

=

12 3

=

4 W

Step 4 - Determine the current through R_eq.

I

=

DV Req

=

12 4

=

3 A

Now we need to expand the circuit back to the original four resistors, and determine the current through, and potential difference across, each one as we go.

Step 1 - R_eq represents R₁ and R₂₃₄ in parallel. Resistors in parallel have the same potential difference but split the current. The potential difference across each is 12 volts. The current is:

I4

=

DV R4

=

12 6

=

2 A

I234

=

DV R234

=

12 12

=

1 A

Step 2 - R₂₃₄ represents R₄ and R₂₃ in series. Devices in series have the same current through them (whatever current their series combination has), so they each have 1A.

DV₄ = I * R₄ = 1 * 3 = 3 V.

DV₂₃ = I * R₂₃ = 1 * 9 = 9 V.

These add to the potential difference across the series combination.

Step 3 - R₂₃ represents R₂ and R₃ in parallel. The potential difference in both cases is 9 V.

I2

=

DV R2

=

9 36

=

0.25 A

I3

=

DV R3

=

9 12

=

0.75 A

Step 4 - A good way to check for consistency is to label the potential at different points. Pick some point as a reference (say, 0 V at the negative terminal of the battery) and label other points relative to that. Check that the potential differences across the resistors are consistent with these potential values.