Internal resistance in networks

Last time

Before we start: last time, in E.m.f., internal resistance and V-I graphs, you learned these points. Here they are again on one card. Keep them in mind, because today builds on them.

A long chain of reasoning

The hardest circuit questions are not hard physics. They are long chains. Find a current here, a p.d. there, feed the answer into the next step. The skill is knowing what is the same, and what adds, at every stage.

What stays the same, what adds

In series, the current is the same through every component, and the p.d.s add to the total. In parallel, the p.d. is the same across every branch, and the currents add up in the main wire. The battery's internal resistance is in series with the whole external circuit. So the battery current flows through r, and the terminal p.d. is E minus I r.

Power and charge together

A total charge of six hundred and fifty coulombs moves through resistor P in five hundred and forty seconds. The current is charge over time: one point two amperes. P dissipates four thousand eight hundred joules. The p.d. is energy per unit charge: four thousand eight hundred over six fifty: seven point four volts. Resistor Q carries four point five amperes, so the battery supplies five point seven amperes in all. For a nine volt battery, the lost volts are nine minus seven point four, and r is that over five point seven: zero point two eight ohms.