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Lesson: Chapter 19 — Current Electricity

19.7. Resistors in parallel 4 of 4

Mixed networks

Both rules in one circuit. A network with series and parallel parts is solved by collapsing it a piece at a time until one resistor is left.

Question 1

The problem

Ruwan needs two resistors, one of 3 Ω and one of 40 Ω. But all he can find are 20 Ω and 9 Ω resistors.

  1. Explain briefly how to make a resistance system of 3 Ω using the resistors above.
  2. Draw, using symbols, a circuit diagram of a 40 Ω resistor system made with those resistors.

Question 2

The problem

The figure shows a circuit built from three bulbs of different filament resistances, with a potential difference of 12 V supplied across its ends. The resistance of the connecting wires may be neglected.

Figure to be added

Figure for exercise 19.4 question 2 — bulb B1 of 20 ohms between points P and Q; between Q and R, bulb B2 of 6 ohms with ammeter A1 on one branch and bulb B3 of 12 ohms with ammeter A2 on the other; ammeter A3 in the main line; and a 12 V supply across P and R.

The parts

  1. What is the equivalent resistance between Q and R?
  2. Find the equivalent resistance between P and R.
  3. Which ammeter's reading shows the total current flowing through the circuit?
  4. Find the total current flowing through the circuit.
  5. What is the potential difference between P and Q?
  6. Find the potential difference between Q and R.
  7. What is the current flowing through bulb B1?
  8. If bulb B2 burns out, calculate the current flowing through the circuit.

How to attack question 2

Work from the inside out. The two bulbs between Q and R are the only part that is purely parallel, so that is where to start.

  • Collapse B2 and B3 into a single equivalent resistance between Q and R using the reciprocal rule.
  • That result is now in series with B1, so add it to 20 Ω to get the resistance between P and R.
  • With one resistance and 12 V, Ohm's law gives the total current — and from there you can work back out to each branch.

Part 8 changes the circuit, not just the numbers

If B2 burns out its branch is broken, so no current flows through it at all. What is left is B1 in series with B3 alone — not the parallel pair with one value changed. Redraw before calculating.