Question
In a circuit, a battery with internal resistance is connected to a parallel combination of and , in series with a resistor. Find (a) the current drawn from the battery, (b) the potential difference across the parallel combination, and (c) the power dissipated in the resistor.
Solution — Step by Step
For and in parallel:
External resistance: . Add internal resistance:
Battery current:
The whole flows through :
Both branches see :
Power:
Why This Works
The systematic approach for any DC network: simplify in stages, find the main current, then “walk back” to compute branch currents and individual voltages. Internal resistance just adds in series with the external network — once you fold it in, Ohm’s law gives the battery current directly.
The key identity that makes parallel branches easy: branches in parallel share the same voltage. So once we know , every branch current follows from .
Alternative Method
Current divider rule: in two parallel resistors, the current through one equals the total current times the opposite resistance over the sum:
Same answer in one step — useful for fast multiple-choice problems where you only need one branch current.
Sanity check: total current entering the parallel block must equal sum of branch currents. . Matches the main current. Always do this check before circling your answer.
Common Mistake
Students often forget to include internal resistance and divide by external resistance only — getting , off by . The phrase “EMF” tells you to use total resistance including ; the phrase “terminal voltage” tells you to use external resistance only.
The other classic slip: assuming the parallel branches carry equal current. They share voltage, not current. The lower-resistance branch carries more current (Ohm’s law). Here the branch carries twice the current of the branch.
Final answer: , , .