H2 Physics 16 - Circuits: Potential Divider with Internal Resistance

Step 1 - State the problem

A battery of e.m.f. twelve volts and internal resistance two ohms is connected to two resistors in series: R one equals four ohms and R two equals six ohms.

Step 1 - State the problem

We want to find the current in the circuit, the terminal voltage of the battery, and the potential difference across each resistor.

Step 2 - Find the current using Kirchhoff's voltage law

By Kirchhoff's voltage law, the e.m.f. equals the sum of the p.d.s around the loop.

Step 2 - Find the current using Kirchhoff's voltage law

E equals I times the total resistance, which is R one plus R two plus the internal resistance r.

Step 2 - Find the current using Kirchhoff's voltage law

So I equals twelve divided by four plus six plus two, which is twelve divided by twelve.

Step 3 - Find the terminal voltage

The terminal voltage V terminal equals the e.m.f. minus the voltage drop across the internal resistance.

Step 3 - Find the terminal voltage

V terminal equals twelve minus one point zero times two, which gives ten volts.

Step 3 - Find the terminal voltage

This is the voltage available to the external circuit.

Step 4 - Find the p.d. across each resistor

The p.d. across R one is V one equals I R one equals one point zero times four equals four volts.

Step 4 - Find the p.d. across each resistor

The p.d. across R two is V two equals I R two equals one point zero times six equals six volts.

Step 4 - Find the p.d. across each resistor

As a check, V one plus V two equals four plus six equals ten volts, which matches the terminal voltage.

Step 5 - Power dissipated and efficiency

The total power from the battery is E times I equals twelve watts.

Step 5 - Power dissipated and efficiency

Power wasted in internal resistance is I squared r equals one times two equals two watts.

Step 5 - Power dissipated and efficiency

Power delivered to the external resistors is ten watts, so the efficiency is ten over twelve, about eighty-three percent.

Step 5 - Power dissipated and efficiency

In exam problems, always check that the lost volts across the internal resistance plus the p.d.s across external components add up to the e.m.f. This is a quick error check.