Consider the circuit shown in A and find the
Looking for parallel or series combinations to reduce to an effective resistance! Must start at the parallel combination of 20k and 5k resistors; it is replaced with its effective resistance of 4k [1/R_{par} = 1/20 + 1/5]. Blue resistors will represent effective resistance, black resistors are real.
The effective resistance of 4k is in series with the actual resistance of 8k, leading to replacement by its effective resistance of 12k. [R_{ser} = 4k + 8k]
Now there is a parallel combination of 12k and 6k resistors; it is replaced with its effective resistance of 4k [1/R_{par} = 1/12 + 1/6].
Finally, the equivalent resistance for the entire circuit is 9k. [R_{ser} = 4k + 5k]. Ohm's law quickly yields a current through the battery of 1 mA. [i = V/R = 9 V/9 k = 1 mA]
The real 5k resistor and the effective 4k resistance each have 1 mA of current since they are in series. Thus the 4k resistance has 4V of voltage difference across it (by Ohm's law).
Breaking the 4k resistance into its component parts (in parallel), we find that 2/3 mA of current flows in the 6k resistor and 1/3 mA flows in the effective resistance of 12k.
Breaking the 12k resistance into its component parts (in series), we find that there is 8/3 V across the 8k resistor and 4/3 V across the effective resistance of 4k.
Finally, breaking the 4k resistance into its component parts (in parallel), we find that 1/15 mA of current flows in the 20k resistor and 4/15 mA flows in the 5k resistor.
In summary,
1. Current through the battery?
2. Current through the 8k resistor? 3. Voltage difference across the 20k resistor? 4. Rate of energy dissipated by the 6k resistor? |
1 mA
1/3 mA 4/3 V P = (2/3 mA) x (4 V) = 8/3 mW |