Here’s How to Crack It
Assume that the ammeter is ideal; it has negligible resistance and doesn’t alter the
current that it’s trying to measure. Similarly, assume that the voltmeter has an
extremely high resistance, so it draws negligible current away from the circuit.
Our first goal is to find the equivalent resistance in the circuit. The 600 Ω and 300
Ω resistors are in parallel; they’re equivalent to a single 200 Ω resistor. This is in
series with the battery’s internal resistance, r, and R 3. The overall equivalent
resistance is therefore Requiv = 50 Ω + 200 Ω + 150 Ω = 400 Ω, so the current
supplied by the battery is I = = 6A. At the junction marked J,